In part (a), the combined area under the standard normal curve is approximately 0.0026. In part (b), the combined area is approximately 0.0686.
(a) To find the combined area under the normal curve, we need to calculate the area to the left of z = -3 and the area to the right of z = 3 separately.
Using a standard normal distribution table or a statistical software, we can find the area to the left of z = -3 is approximately 0.0013. Similarly, the area to the right of z = 3 is also approximately 0.0013.
Now, to find the combined area, we can add these two areas together:
Combined area = 0.0013 + 0.0013 = 0.0026
Therefore, the combined area under the standard normal curve in part (a) is approximately 0.0026.
(b) Similar to part (a), we need to calculate the area to the left of z = -1.53 and the area to the right of z = 2.53 separately.
Using a standard normal distribution table or a statistical software, we can find the area to the left of z = -1.53 is approximately 0.0630. The area to the right of z = 2.53 is approximately 0.0056.
To find the combined area, we can add these two areas together:
Combined area = 0.0630 + 0.0056 = 0.0686
Therefore, the combined area under the standard normal curve in part (b) is approximately 0.0686.
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Complete question :
Part 2 of 3 Determine the total area under the standard normal curve in parts (a) through (c) below. CHFF (a) Find the area under the normal curve to the left of z=-3 plus the area under the normal curve to the right of z = 3. The combined area is 0.0028 (Round to four decimal places as needed.) (b) Find the area under the normal curve to the left of z= -1.53 plus the area under the normal curve to the right of z=2.53. The combined area is (Round to four decimal places as needed.)
Answer the following questions. Explain your answer further.
1. Is the hydrostatic pressure the same along any constant horizontal line?
2. What is the effect of temperature on the hydrostatic pressure?
1. The hydrostatic pressure varies along different horizontal lines due to differences in depth.
2. The temperature, although capable of affecting the density of the fluid, does not influence hydrostatic pressure.
The hydrostatic pressure is not the same along any constant horizontal line. The pressure at a specific point depends on its depth within the fluid. If two points have the same depth, they will experience the same pressure. However, if two points are located at different horizontal lines but have the same depth, their pressures will also be equal. On the other hand, when the depths of two points differ, the pressure at the deeper point will be greater than that at the shallower point.
In terms of temperature, its direct effect on hydrostatic pressure is negligible. Hydrostatic pressure is primarily determined by the depth of the point and the density of the fluid. Assuming a constant fluid density, the temperature has no immediate impact on hydrostatic pressure. However, the temperature can indirectly influence pressure by altering the fluid's density. When the temperature of the fluid increases, its density decreases. Consequently, this reduction in density leads to a decrease in hydrostatic pressure since pressure is directly proportional to fluid density. Nevertheless, the change in hydrostatic pressure resulting from temperature fluctuations is generally small, and it can usually be disregarded in most practical applications.
In summary, the hydrostatic pressure varies along different horizontal lines due to differences in depth. The temperature, although capable of affecting the density of the fluid, has only a minor influence on hydrostatic pressure and is typically not a significant consideration in practical scenarios.
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What is the percentage of nitrogen, by mass, in calcium nitrate?.
Answer:
Nitrogen = 17.07%
Explanation:
We are asked to calculate the percentage by mass of nitrogen in Ca(NO3)2.
The molar mass of calcium nitrate is 164,088 g/mol.
In this 164.088 g/mol there are 2 nitrogen atoms.
Molar mass of nitrogen = 14,0067 g/mol
% Nitrogen = molar mass of N/total molar mass of the compound
% Nitrogen =[(2 x 14.0067)/164.088]*100
% Nitrogen = 17.07%
Please mark me as brainliest if you found this answer helpful
Help me please I need to write the number of atoms
Please help me no links
if good answer you get brainliest
Answer:
plutonic rocks are formed when magma cools and solidifies underground This shows us if the rock is plutonic or volcanic. When magma cools underground, it cools super slow and when lava cools above ground, it cools faster. When magma and lava cool, mineral crystals start to form in the molten rock.
Explanation:
i hope this is correct
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
A) .57 grams
B) .70 grams
C) 2.25 grams
D) .75 grams
Answer:
D. 0.75 grams
Explanation:
The data given on the iridium 182 are;
The half life of the iridium 182, \(t_{(1/2)}\) = 15 years
The mass of the sample of iridium, N₀ = 3 grams
The amount left, N(t) after two half lives is given as follows;
\(N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }\)
For two half lives, t = 2 × \(t_{(1/2)}\)
∴ t = 2 × 15 = 30
\(\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2\)
\(\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75\)
∴ The amount left, N(t) = 0.75 grams
how much more kinetic energy does a 6 kg bowling ball have when it is rolling at 16 mph then when it is rolling at 14 mph
The difference in kinetic energy if a 6 kg bowling ball rolling at 16 mph and when it is rolling at 14 mph is 180J.
How to calculate kinetic energy?Kinetic energy of an object can be calculated using the following formula;
K.E = ½mv²
Where;
K.E = kinetic energym = massv = velocityAccording to this question, a 6 kg bowling ball is rolling at 16 mph and 14 mph respectively.
K.E = (½ × 6 × 16²) - (½ × 6 × 14²)
K.E = 768 - 588
∆K.E = 180J
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Al + NaOH + H₂O
NaOH + Al₂O3+H₂
Answer:
(o) Al + NaOH + H2O -NaAlO2 + H2.
When aluminum oxide Al 2 O 3 reacts with sodium hydroxide solution, it forms sodium aluminate and water H 2 O . This reaction takes place at a temperature of 900 - 1100 ° C
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide:
4AI (s) + 302 (g) → 2Al2O3 (s)
The maximum amount of Al2O3 that can be produced from 5.0 g of Al and 5.0 g of O2 is.
.g.
Answer:
5g Al: 9.45g
5g O2: 10.62g
Explanation:
view attachment for worked out problem
Explain or show how you could calculate the mass of an object if you know its volume and density.
Answer:
yea I'm also looking for this question to .
Helppppp plz ASAP DON’T GUESS
Explanation:
A B C D isme koyi ans hain doond lo
what do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 752 torr ?
8.52 m is the height of a barometer column based on 1-iodododecane.
What is a barometer?A barometer is described as a scientific instrument that is used to measure air pressure in a certain environment.
For the given question we will use the below formula:
P = dgh, where
g = gravitational force = 9.8 m/s²
h = height
First we calculate the height of the barometer column for the mercury:
Density of mercury = 13.6g/ml (given)
Given pressure = 752 torr =100258.144 N/m²
Height of barometer for mercury = 100258.144 / (13.6×9.8) = 752.23= 0.752
Now we calculate the height of barometer by using the below formula:
d₁h₁ = d₂h₂, where
d₁ = density of 1-iodododecane = 1.20g/mol (given)
h₁ = to find?
d₂, h₂ = density & height with respect to mercury
On putting all values in the above equation we get,
h₁ = 13.6×0.752 / 1.20 = 8.52m
Hence, 8.52m is the height of barometer.
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Complete question:
The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 752 torr ?
What might happen if you mixed a strong acid with an equally strong base?
Question 4 options:
You would wind up with a pH neutral salt and water
The base would destroy the acid.
You would see an explosive chemical reaction.
The acid would destroy the base.
Answer:
You would wind up with a pH neutral salt and water
Explanation:
When they are equally strong they will both neutralize each other, and the acidic and basic properties are no longer there.
How many moles of O2 would there be if I had 4 moles of Fe
4 moles of Fe react with 3 moles of oxygen. Therefore, the mole ratio of iron to iron oxide in this process is 4:2.
What is oxygen ?The chemical element with the atomic number 8 and symbol O is called oxygen. It belongs to the periodic table's halogen group, is a very reactive nonmetal, and an oxidizing agent that easily produces oxides with most elements as well as other compounds.
The non-metallic element oxygen occurs naturally as a molecule. Two oxygen atoms that are tightly bound together make up each molecule. Oxygen is in a gaseous form at ambient temperature due to its low melting and boiling temperatures.
According to scientists, the oceans produce between 50 and 80 percent of the oxygen used on Earth. Oceanic plankton, which includes floating plants, algae, and certain bacteria that can photosynthesize, is the main source of this production.
Thus, 4 moles of Fe react with 3 moles of oxygen.
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A gas has a volume of 0.60L at 161K. Once heated, the same gas now has a volume of 14.1L at 279K and 2.44atm. What was the original pressure of the gas?
Answer:
31.08 atm.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 0.60 L
Initial temperature (T₁) = 161 K
Final volume (V₂) = 14.1 L
Final temperature (T₂) = 297 K
Final pressure (P₂) = 2.44 atm
Initial pressure (P₁) =?
The initial pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
P₁ × 0.6 / 161 = 2.44 × 14.1 / 297
P₁ × 0.6 / 161 = 34.404 / 297
Cross multiply
P₁ × 0.6 × 297 = 161 × 34.404
P₁ × 178.2 = 5539.044
Divide both side by 178.2
P₁ = 5539.044 / 178.2
P₁ = 31.08 atm
Thus, the initial pressure of the gas was 31.08 atm.
Question 34
The fissionable fuel in all US nuclear reactors is:
a. Plutonium
b. Thorium
c. Uranium
d. tritium
The use of thorium and tritium in nuclear energy production. Thorium is a naturally occurring radioactive metal that can be used as a fuel in nuclear reactors. It is considered to be a safer and more abundant alternative to uranium, as it produces less radioactive waste and is more readily available. Answer is b
Thorium is not a fissile material and must be converted into uranium-233 through a process called breeding in order to be used as fuel.
Tritium, on the other hand, is a radioactive isotope of hydrogen that can be used in fusion reactions to produce energy. Fusion reactions involve combining atomic nuclei to release energy, and tritium is one of the fuels used in this process. However, tritium is difficult to produce and must be constantly replenished in order to sustain a fusion reaction.
Both thorium and tritium have the potential to provide clean and sustainable sources of energy. Further research and development are needed to make these technologies commercially viable and safe for widespread use.
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Question 11
Which formula represents a hydrocarbon?
C₂H6
C₂H5OH
C₂H5Cl
C₂H6O
Answer:
C₂H6
Explanation:
Among the given options, the formula A) C₂H6 represents a hydrocarbon (specifically, ethane). Option A
A hydrocarbon is a compound that consists of only carbon and hydrogen atoms. It is important to identify the formula that represents a hydrocarbon among the given options:
A) C₂H6: This formula represents ethane, which is a hydrocarbon. Ethane consists of two carbon atoms bonded together with single bonds and six hydrogen atoms.
B) C₂H5OH: This formula represents ethanol, which is not a hydrocarbon. Ethanol contains a hydroxyl group (-OH), indicating the presence of oxygen in addition to carbon and hydrogen atoms. It is an alcohol, not a hydrocarbon.
C) C₂H5Cl: This formula represents ethyl chloride, which is not a hydrocarbon. Ethyl chloride contains a chlorine atom (Cl) in addition to carbon and hydrogen atoms. It is a haloalkane, not a hydrocarbon.
D) C₂H6O: This formula represents ethanol, which, as mentioned before, is not a hydrocarbon. Ethanol contains an oxygen atom (O) in addition to carbon and hydrogen atoms. It is an alcohol, not a hydrocarbon.
Among the given options, the formula A) C₂H6 represents a hydrocarbon (specifically, ethane). It consists only of carbon and hydrogen atoms, making it a suitable representation of a hydrocarbon.
In summary, the formula C₂H6 (option A) represents a hydrocarbon, while the other options contain additional elements (oxygen or chlorine) that make them non-hydrocarbon compounds. Option A
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describe any four factors that bring out a chemical change.
Answer:
Physical state of reactants
Temperature
Concentration of reactants
Presence of a catalyst
Explanation:
The larger the surface area of a reactant, the faster the rate of the reaction. If the reactant is in powdered form, the reaction takes place faster than when the reactant is in solid or block form. This is due to the fact that the latter has larger surface area
Increase in temperature causes reactant molecules to gain kinetic energy thus increasing the rate if reaction
Pls help I need it ASAP!!!
Answer:
B
Explanation:
because water has the always has the same volume but it can change shape of the glass or container that it is put in
A lump of zinc is tossed into a beaker of 500L of 14M hydrochloric acid. this reaction produces Hydrogen Gas and zinc (II) chloride. If the hydrogen gas is combusted and produces 645L of water vapor at 400 kelvin and 1.75 atm, what is the mass of the zinc?
If the hydrogen gas is combusted and produces 645L of water vapor at 400 kelvin and 1.75 atm, 2796.96 g mass of the zinc is produced .
Using the ideal gas law equation:
PV = nRT
n = (PV) / (RT)
= (1.75 atm * 645 L) / (0.0821 atm·L/(mol·K) * 400 K)
= 42.71 moles
the balanced equation for the reaction between zinc and hydrochloric acid:
Zn + 2HCl -> \(ZnCl_{2}\) + \(H_{2}\)
1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the moles of zinc are also 42.71.
The molar mass of zinc is 65.38 g/mol.
Mass of zinc = moles of zinc * molar mass of zinc
= 42.71 moles * 65.38 g/mol
= 2796.96 g
Therefore, the mass of the zinc is 2796.96 grams.
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Determine physiological temperature, 98.6 F in degree C
Answer:
37
Explanation:
( 98.6 - 32 ) × 5(100c) ÷ 9(180f) = 37
If 12.3 mol HCl are produced in this reaction, how many grams of sodium sulfate are produced?
ANSWER
The mass of Na2SO4 is 874g
EXPLANATION
Given that;
The number of moles of HCl is 12.3 mol
Follow the steps below to find the mass of sodium sulfate produced
Step 1; Write a balanced equation for the reaction
\(\text{ 2NaCl + H}_2SO_4\text{ }\rightarrow\text{ 2HCl + Na}_2SO_4\)In the reaction above, 2 moles of NaCl react with 1 mole of H2SO4 to give 2 moles of HCl and 1 mole of Na2SO4
Let the number of moles of Na2SO4 be x
\(\begin{gathered} \text{ 2 moles HCl }\rightarrow\text{ 1 mole Na}_2SO_4 \\ \text{ 12.3 moles HCl }\rightarrow\text{ x moles Na}_2SO_4 \\ \text{ Cross multiply} \\ \text{ 2 moles HCl }\times\text{ x moles Na}_2SO_4\text{ }=\text{ 1 mole Na}_2SO_4\text{ }\times\text{ 12.3 mole HCl} \\ \text{ Isolate x} \\ \text{ }\times\text{ }=\text{ }\frac{1\text{ mole Na}_2SO_4\times12.3mol\cancel{HCl}}{2moles\cancel{HCl}} \\ \text{ } \\ \text{ x = }\frac{1\text{ }\times\text{ 12.3}}{2} \\ \text{ x = 6.15 moles} \end{gathered}\)The number of moles of Na2SO4 is 6.15 moles
Step 3; Find the mass of Na2SO4 using the below formula
\(\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}\)Recall, that the molar mass of Na2SO4 is 142.04 g/mol
\(\begin{gathered} \text{ mass = mole }\times\text{ molar mass} \\ \text{ mass = 6.15 }\times\text{ 142.04} \\ \text{ mass = 873.546} \\ \text{ mass = 874g Na}_2SO_4 \end{gathered}\)Therefore, the mass of Na2SO4 is 874g
Chloroform is a commonly used anesthetic with a density of 1.483 g/mL. Determine the volume of chloroform needed to deliver a 9.37 g sample of the anesthetic
Answer:
The correct answer is 6.32 ml
Explanation:
We have the following data:
Mass : m = 9.37 g
Density : d= 1.483 g/mL
The density a substance is equal to the mass of the substance divided into its volume. So, we calculate the volume (V) of chloroform from the mass and density as follows:
\(d=\frac{m}{V}\) ⇒ \(V = \frac{m}{d} = \frac{9.37 g}{1.483 g/ml} = 6.32 ml\)
A hydrocarbon contains 7.7% by mass of hydrogen and 92.3 by mass of carbon and the relative molecular mass is 78 derive the empirical formula of the compound and the molecular formula
Answer:
molecular mass = 78
Empirical Formula is
C= 92.3/12= 7.69
H= 7.68/1= 7.68
therefore Empirical formula is
CH
molecular formula is
(CH)n= 78
(12+1)n=78
n=78/13
n=6
molecular formula is
C6H6
Express the dosage using the ratio format you prefer. (Use mg for milligrams and mL for an injectable solution that contains 250mg in each 0.6 mL 3. [-/3 Points] CURRENMEDMATH11 12.3.002. EP. Consider the following. A 40mg in 2.5 mL solution will be used to prepare a 26mg dosage. Calculate the dosage using ratio and proportion. Express your final answer in mL to the
40mg
mL
=
X mL
26mg
40x
X
=
=
mL
[-/1 Points] CURRENMEDMATH11 12.3.004. Calculate the dosage (in milliliters). Express your answer to the nearest tenth. Assess y A 36mg per 2 mL strength solution is used to prepare 22mg. mL
The dosage of 26mg can be prepared using approximately 1.625 mL of the 40mg in 2.5 mL solution.
The dosage of 22mg can be prepared using approximately 1.222 mL of the 36mg per 2 mL strength solution.
To calculate the dosage using ratio and proportion, we can set up a proportion based on the strength of the solution.
40mg in 2.5 mL solution will be used to prepare a 26mg dosage.
Let X represent the mL of the solution needed to prepare the 26mg dosage.
We can set up the proportion as follows:
40mg/2.5mL = 26mg/X mL
Cross-multiplying and solving for X, we have:
40mg * X mL = 2.5mL * 26mg
40X = 65
X = 65/40
X ≈ 1.625 mL
For the second question:
36mg per 2 mL strength solution is used to prepare 22mg.
Let Y represent the mL of the solution needed to prepare the 22mg dosage.
We can set up the proportion as follows:
36mg/2mL = 22mg/Y mL
Cross-multiplying and solving for Y, we have:
36mg * Y mL = 2mL * 22mg
36Y = 44
Y = 44/36
Y ≈ 1.222 mL
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Determine the mass of zirconium_______________, silicon__________________, and oxygen________________, found in 0.3384 mol of zircon, ZrSiO4, a semiprecious stone. Your answer must have the correct number of significant figures.Enter the number only, with no units.
1) Determine the mass of zirconium
Convert moles of zircon into moles of zirconium
\(\text{molesofZr}=0.3384molesofZrSiO_4\cdot\frac{1\text{molofZr}}{1molofZrSiO_4}=0.3384\text{molesofZr}\)Convert moles of zirconium into mass of zirconium (g)
\(\text{gramsofZr}=0.3384\text{molesZr}\cdot\frac{91.224\text{ g Zr}}{1\text{molofZr}}=30.87\text{ g Zr}\)2) Determine the mass of Silicon
Convert moles of zircon into moles of silicon
\(\text{molesofSi}=0.3384molesofZrSiO_4\cdot\frac{1\text{molofSi}}{1molofZrSiO_4}=0.3384\text{molesofSi}\)Convert moles of silicon into mass of silicon (g)
\(\text{gramsofSi}=0.3384\text{molesofSi}\cdot\frac{28.085\text{ g of Si}}{1\text{molofSi}}=9.504\text{ g Si}\)3) Determine the mass of Oxygen
Convert moles of zircon into moles of oxygen
\(\text{molesofO}=0.3384molesofZrSiO_4\cdot\frac{4\text{molofO}}{1molofZrSiO_4}=1.3536\text{molesofO}\)Convert moles of oxygen into mass of oxygen
\(\text{gramsofO}=1.3536\text{molesofO}\cdot\frac{15.999\text{ g O}}{1\text{mol of O}}=21.66gO\)The mass of zirconium is 30.87 g
The mass of silicon is 9.504 g
The mass of oxygen is 21.66 g
why are we always at the center of a rainbow? and why can we see a rainbow?
Answer:
the sun is directly behind your head
Explanation:
The reason why we can see a rainbow is caused by sunlight and atmospheric conditions
The molar mass of an unknown compound is 560 g. A sample of the compound consists of 0.900 g of carbon, 0.0751 g of hydrogen, 0.175 g of nitrogen, and 0.600 g of oxygen. What is the the molecular formula of this compound. O C24H₂4N4012 O C24H22N3013 C12H14N₃O5 O C24H₂8 N010
The molecular formula of the compound is C12H14N₃O5.
To determine the molecular formula of the compound, we need to calculate the empirical formula first, which represents the simplest whole-number ratio of atoms in the compound.
Given the masses of carbon, hydrogen, nitrogen, and oxygen in the sample, we can calculate the moles of each element using their molar masses:
Moles of C = 0.900 g / 12.01 g/mol = 0.0749 mol
Moles of H = 0.0751 g / 1.008 g/mol = 0.0745 mol
Moles of N = 0.175 g / 14.01 g/mol = 0.0125 mol
Moles of O = 0.600 g / 16.00 g/mol = 0.0375 mol
Next, we need to find the simplest ratio of the moles by dividing each value by the smallest value:
Moles of C / 0.0125 = 5.992
Moles of H / 0.0125 = 5.960
Moles of N / 0.0125 = 1.000
Moles of O / 0.0125 = 3.000
Rounding these ratios to the nearest whole number, we get a ratio of 6:6:1:3, which corresponds to the empirical formula C6H6N1O3.
Finally, to determine the molecular formula, we divide the given molar mass of the compound (560 g) by the molar mass of the empirical formula (C6H6N1O3):
560 g / (6 * 12.01 g/mol + 6 * 1.008 g/mol + 1 * 14.01 g/mol + 3 * 16.00 g/mol) ≈ 560 g / 194.19 g/mol ≈ 2.88
Since the result is close to 3, we can multiply the empirical formula by 3 to obtain the molecular formula: C6H6N1O3 * 3 = C18H18N3O9.
However, none of the options provided match the calculated molecular formula.
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The molecular formula for the molar mass of the unknown compound is 560g that consists of 0.900 g of carbon, 0.0751 g of hydrogen, 0.175 g of nitrogen, and 0.600 g of oxygen is C₂₄H₂₄N₄O₁₂ (Option A).
To determine the empirical formula, which involves converting the sample into moles. The moles of each element in the compound are calculated using their respective atomic masses.
C = 0.900/12.01 = 0.0749 H = 0.0751/1.01 = 0.0745 N = 0.175/14.01 = 0.0125 O = 0.600/16.00 = 0.0375The smallest number of moles is 0.0125 moles of nitrogen, which is the limiting reagent. As a result, the empirical formula is:
N = 0.0125/0.0125 = 1C = 0.0749/0.0125 = 6H = 0.0745/0.0125 = 6O = 0.0375/0.0125 = 3Therefore, the empirical formula is C₆H₆NO₃.
The empirical formula mass can be calculated by adding the molar masses of each element:
C = 6(12.01) = 72.06 H = 6(1.01) = 6.06 N = 1(14.01) = 14.01 O = 3(16.00) = 48.00Total mass = 140.13
The molecular formula can be determined by comparing the empirical formula mass and the given molar mass. The molecular formula is the same as the empirical formula when the two values are equal. The ratio of the molecular formula mass to the empirical formula mass is equal to the integer value of n (number of empirical formula units):
n = molar mass/empirical formula mass
n = 560/140.13
n = 4
Therefore, the molecular formula is four times the empirical formula: C₂₄H₂₄N₄O₁₂ (Option A).
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a sample of gas held at constant volume is raised from 34.0 atm to 5.90 atm. if the final temperature of the sample is 340.0 K, what was its initial temperature
Answer
To answer this question we need to use the following equation
P1/T1 = P2/T2
We can rearrange to make T1 subject of the formula as shown
T1 = (P1 x T2)/ P2
= (3.40 x 340)/ 5.90
= 195.9 K
Which solution will boil at the highest temperature? *
1 mole of sugar in 500 g of water
1 mole of sugar in 1,000 g of water
2 moles of sugar in 500 g of water
2 moles of sugar in 1,000 g of water
Answer:
2 moles of sugar in 1,000g of water
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