is the information provided enought to determine the exact position of m? if not, what other solutions would you need to test?

Answer:

275g

Explanation:

Depending on the molar mass you are given, you can use that to solve this.

(I'm going based on my science class' molar mass of sulphur being 32.07g/mol)

Starting off, the formula for finding moles is

n=m/M (moles = mass / molar mass)

We can manipulate this equation to solve for mass.

m=Mn

now fill in what we now.

m = 32.07*8.56

mass = 274.5192

Now round for significant digits (if you are needed to do)

mass = 275g

a) The temperature of the system when thermal equilibrium is reached will be 0°C.

b) 30 g of ice will be present when thermal equilibrium is reached.

mass of ice (m1) = 50.0 g

Temperature of ice (T1) = -22.0°C

Mass of water (m2) = 120.0 g

Temperature of water (T2) = 7.0°C

The energy required to melt the ice is given by the equation,

Q1 = m1 × Lf

Where, Lf is the latent heat of fusion of ice = 334 J/g

Q1 = 50.0 × 334Q1 = 16700 J

The energy required to heat the ice from -22°C to 0°C (Q2) is given by,

Q2 = m1 × c × (0-(-22))

Where, c is the specific heat capacity of ice = 2.06 J/g°C

Q2 = 50.0 × 2.06 × 22Q2 = 2266 J

The energy lost by water (Q3) is given by the equation,

Q3 = m2 × c × (7 - 0)

Where, c is the specific heat capacity of water = 4.184 J/g°C

Q3 = 120 × 4.184 × 7Q3 = 35244.48 J

Total energy gained (Q4) by ice and water is equal to the energy lost by the water.

Q4 = Q1 + Q2

Q4 = 16700 + 2266

Q4 = 18966 J

18966 = Q3 = m2 × c × (7-0)

18966 = 120 × 4.184 × 7

m2 = 18966/(120 × 4.184 × 7)

m2 = 3.03 g

At equilibrium, the mass of the remaining ice (m3) can be calculated as follows,

Q1 + Q2 = m3 × Lf + m3 × c × (0 - 0°C)

16700 + 2266 = m3 × 334 + m3 × 2.06 × (0 - (-22))

m3 = 30 g

Therefore, the temperature of the system when thermal equilibrium is reached will be 0°C, and the mass of the ice remaining at equilibrium will be 30 g.

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The balanced nuclear equation for the synthesis of 289 Uup (element 115) is as follows:
243 Am + 48 Ca → 289 Uup + 4 n
This reaction involves firing a beam of 48 Ca ions at 243 Am. The collision of the two nuclei results in the creation of a superheavy element, 289 Uup. In the process, two neutrons are also ejected. The reaction is balanced with the lowest possible coefficients, indicating that for every one 243 Am and 48 Ca that react, one 289 Uup and four neutrons are produced. The synthesis of superheavy elements through nuclear reactions such as this one is an area of ongoing research in nuclear physics.

To write a balanced nuclear equation for the synthesis of the superheavy element 289Uup (element 115) using the given terms, we can follow these steps:
1. Identify the initial reactants: 48Ca ions and 243Am.
2. Recognize that two neutrons are ejected during the reaction.
3. Determine the resulting product, which is 289Uup.
The balanced nuclear equation can be written as:
48Ca + 243Am -> 289Uup + 2n
This equation indicates that when a beam of 48Ca ions is fired at 243Am, it results in the synthesis of the superheavy element 289Uup, along with the ejection of two neutrons.

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Explanation:

the answer for the above

is b)0.019

The empirical formulas of four binary ionic compounds formed from the given ions are:

The empirical formula of a compound represents the simplest ratio of the elements present in it. In this case, we have three ions: Mg\(_{2}\)+, Au\(_{3}\)+, and I-. To form a binary ionic compound, the positive and negative charges of the ions need to balance.

For the first compound, magnesium iodide (MgI\(_{2}\)), two iodide ions (I-) are needed to balance the charge of one magnesium ion (Mg\(_{2}\)+).

For the second compound, gold(III) oxide (Au\(_{2}\)O\(_{3}\)), three oxide ions (O\(_{2}\)-) are required to balance the charge of two gold ions (Au\(_{3}\)+).

The third compound is magnesium oxide (MgO), where one oxide ion (O\(_{2}\)-) combines with one magnesium ion (Mg\(_{2}\)+).

Lastly, for gold(I) iodide (AuI), one iodide ion (I-) is required to balance the charge of one gold ion (Au+).

These empirical formulas represent the simplest combination of ions in these ionic compounds.

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Benefits of using micro-scale techniques is that they offer a high level of precision and control in scientific experimentation. By using micro-scale techniques, researchers can manipulate small amounts of materials and samples, allowing them to perform experiments with a greater degree of accuracy and repeatability.

This can be especially useful in fields such as biology and chemistry, where even small variations in experimental conditions can have a significant impact on the results.Benefits of using micro-scale techniques is that they can reduce the cost and time required for experimentation. By using smaller samples and less reagents, researchers can save money on materials and reduce the time required for experiments to be completed. In addition, micro-scale techniques can be more environmentally friendly, as they require less waste and energy to produce.

Benefits of micro-scale techniques could include examples of specific applications in various scientific fields, such as microfluidics for drug discovery or microscale electrophoresis for DNA analysis. It could also discuss how micro-scale techniques are advancing research in areas such as nanotechnology and biomedicine, and how they are helping to solve some of the world's most pressing scientific challenges. Overall, the benefits of using micro-scale techniques are numerous and varied, and they are likely to continue to play an important role in scientific experimentation for years to come.

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Answer:

C. The electrons in each C-F bond will be more attracted to the fluorine atom.

The structure of an ozone molecule is a resonance structure of three oxygen atoms covalently bonded together.

An ozone molecule is a form of oxygen molecule that is composed of three atoms of oxygen covalently bonded to each other.

The formula of the ozone molecule is given as O₃.

An ozone molecule is an important form of oxygen in that it is found in the layer of the atmosphere and acts as a protective layer against harmful ultraviolet radiation from the sun.

The lewis dot structure of ozone shows that an ozone molecule has two resonance structures.

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Answer:

1

Explanation:

mark as brainliest plz

Answer:

The answer is not 1, it's 2.

Explanation:

I chose one and I got it incorrect, unaided eye is obviosly not it. It thenhas to be meter stick.

In a weak acid and strong base titration, the pH at the equivalence point is affected by: the basicity of the salt anion. Hence, the final pH of the equivalence factor will be higher than 7.

At the equivalence point, the mixture of a weak acid and a strong base produces basic salt. This happens because the acid is too weak to donate the hydrogen ions to the solution, so it has an anion of a strong base. Meanwhile, the base is strong enough to accept hydrogen ions from water, so it has a cation of a weak acid. It affects the salt pH to be greater than 7 (basic salt).

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Answer:

d . al of these

Explanation:

all of that are used when you in spaceship

The complete information is as follows:

Formula: A. Al₂(SO₄)₃; B. Al₂(SO₄)₃; C. Al₂(SO₄)₃; D. Ca(NO₃)₂; E. Ca(NO₃)₂

Molar Mass (g/mol): A. 342.15 g/mol; B. 342.15 g/mol; C. 342.15 g/mol; D. 164.09 g/mol; E. 164.09 g/mol

Number of particles: A. 8.34*10²³; B. 4.9110²⁴; C. 1.2010²⁴; D. 2.44*10²³; E. 5.00*10²³

Number of moles; A. 0.014 moles;  B. 0.143 moles; C. 3.50 moles; D. 0.149 moles; E. 0.458 moles

Mass (grams): A. 4.66 g; B. 49.60 g; C. 1190.35 g; D. 42.7 g; E. 75.03 g

The number of particles in a compound is determined using the formula below:

Number of particles = number of moles * 6.02 * 10²³

The number of moles is determined as follows:

Number of moles = mass / molar mass

or

Number of moles = Number of particles / 6.02 * 10²³

The mass is determined as follows:

mass = number of moles * molar mass

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P-aminophenol has a molar mass of 109.13 g/mol. Acetaminophen has a molar mass of 151.17 g/mol. Acetic anhydride has a capacity of 1.1 mL.

The mass of acetaminophen, stated as 0.157g, must be multiplied by the molar mass of acetaminophen to get the theoretical yield. It weighs 151.2g in this case. The theoretical yield thus becomes 0.217g.

A measurement called production yield is obtained by dividing the number of high-quality parts produced by the total number of parts started in production.

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Answer: The reaction that occurs at anode is \(Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-\)

Explanation:

Given : \(E^o_{Zn^{2+}/Zn}=-0.76V\)

\(E^o_{Cu^{2+}/Cu}=+0.34V\)

The substance having highest positive reduction potential will always get reduced and will undergo reduction reaction. Here, copper will undergo reduction reaction will get reduced.

The substance having highest negative reduction potential will always get oxidised and will undergo oxidation reaction. Here, zinc will undergo oxidation reaction will get oxidised.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Oxidation half reaction (anode) : \(Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-\)

Reduction half reaction (cathode) : \(Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)\)

Answer:

chemical energy

Explanation:

just put that

Answer:

chemical energy

Explanation:

it is the light dependent reaction

Answer:

Too much iodine can be harmful if you don't need it. You don't need to take iodine (supplements) if you already have a varied and balanced diet.

Explanation:

Iodine is already contained in table salt so you don't need to add any more.

Answer:

42 L

Explanation:

de los parámetros en la pregunta;

V1 = 358L

T1 = 152 ° C + 273 = 425 K

P1 = 470 mmHg × 1 atm / 760 mmHg = 0.6atm

V2 =?

P2 = 6 atmósferas

T2 = 500 K

P1V1 / T1 = P2V2 / T2

P1V1T2 = P2V2T1

V2 = P1V1T2 / P2T1

V2 = 0,6 × 358 × 500/6 × 425

V2 = 107400/2550

V2 = 42 L

2.50 litres of a 10.0 M solution require the preparation of 25.0 moles of ethylene glycol.

Excellent antifreeze, anti-boil, and anti-corrosive qualities are produced when antifreeze and water are mixed in a 50/50 ratio. The proportion of conventional ethylene glycol to water in severely cold conditions can reach 70% antifreeze, 30% water. The maximum antifreeze to water ratio when using DEX-COOL® is 60/40.

moles = concentration (M) x volume (L)

Given that the desired concentration is 10.0 M and the volume needed is 2.50 L, the setup for calculating the total number of moles of ethylene glycol can be written as:

moles = 10.0 M x 2.50 L

moles = 25.0 mol

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Answer:

The second choice and can i have brainiest

Explanation:

Answer:

Because they have complete electronic configuration

Explanation:

Group 18 elements are known as the noble or inert gases. The elements in this group are non-reactive because they have complete electronic shell configuration.

The number of grams of KClO₃ needed to make 0.402 moles of oxygen gas is 32.8 grams

Stoichiometry refers to the quantitative relationship between the reactants and products of a specific reaction or equation.

According to this question, pottasium chlorate decomposes to form pottasium chloride and oxygen gas as follows:

2KClO₃ → 2KCl + 3O₂

2 moles of pottasium chlorate decomposes to form 3 moles of oxygen gas.

0.402 moles of oxygen gas will be produced as a result of the reaction of 0.268 moles

mass = 0.268 moles × 122.55 g/mol

mass = 32.8 grams.

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If the reactants of a chemical reaction contain three total carbon atoms the products of the chemical equation will have 3 carbon atoms.

Since matter cannot be created or destroyed, the rule of conservation of mass states that no matter is synthesized or degraded during a chemical reaction. It is only transformable.

As a result, atoms are preserved in chemical reactions. This means that the elements that are forming a compound in a specific ratio will reorganize to create new compounds with the same number and kind of atoms as the reactants.

In conclusion, if a molecular equation's reactants contain 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms, its products will also contain 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms.

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The complete question is:

If the reactants of a chemical equation have 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms. In order to follow the conservation of matter, the products of the chemical equation will have

The solution to the equation and value of variable 21(3d - 4) + 100 = 58 is d = 2/3.

Solve the equation 21(3d - 4) + 100 = 58, we can begin by simplifying and isolating the variable:

21(3d - 4) + 100 = 58

Distribute 21 to the terms inside the parentheses:

63d - 84 + 100 = 58

Combine like terms:

63d + 16 = 58

Subtract 16 from both sides:

63d = 42

Divide both sides by 63:

d = 42/63

Simplifying the fraction gives:

d = 2/3

The solution to the equation is d = 2/3.

The solution to the equation 21(3d - 4) + 100 = 58 is d = 2/3. By simplifying the equation, we find that dividing both sides by 63 results in the solution of d = 2/3, which satisfies the original equation.

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A solid produced by a chemical reaction in solution that separates from the solution is called Precipitate.

Precipitate: a solid that forms from a chemical reaction taking place in solution and separates from the solution.

a solid created when a solution undergoes a change, frequently as a result of a chemical reaction or temperature shift that makes a solid less soluble. A precipitate in meteorology is either liquid or solid water (rain, snow, etc.)

The clear liquid remaining above the precipitated or the centrifuged solid phase is also called the 'Supernate' or 'Supernatant'.

Therefore, the chemical reaction in solution that separates from the solution is called precipitate.

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False. In a closed system, matter can not enter or exit that is there is no change in the matter of the system.

Three types of systems exist in nature:

1. Open System: In this system, the matter can interact with the surroundings or matter can enter or exit the system from the surrounding. Similarly, the energy of the system also interacts with its surroundings and can be lost or gained.

For example oceans etc.

2. Closed system: In this system, the matter is unable to interact with the surroundings that are matter can't exit or enter the system. While the energy of the system is able to interact with the surroundings.

For example Earth etc

3. Isolated system: In this system, both matter and energy are unable to interact with the surrounding. There is no exchange between matter and the energy of surroundings.

For example thermos-teel bottles etc.

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The volume at which the pH can be used to determine the pKa of the acid is 7.40 ml if the equivalence point occurred at 14.81 ml of added sodium hydroxide.

The weak acid and the conjugate base will have an equal number of moles at the half-equivalence point as the number of moles of sodium hydroxide will neutralize half of the number of moles of the weak acid which will then produce more conjugate base.

The chemical reaction can be given as;

HA = (H^+) + A^-

Calculation of pKa:

pH = pKa + log[A^-]/[HA]

At half equivalence point, pH = pKa

As, ka = 10^-pka

Therefore, ka = 10^-pH

Volume is given which is 14.81 ml. Therefore, the volume that is used to determine the pKa of the acid can be calculated as follows;

14.81/2 = 7.40 ml

Therefore the volume used to determine the pKa of the acid is calculated to be 7.40 ml

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Okay, let's think through this step-by-step:

1) The initial equilibrium lies on the right side, favoring the products (H2 and O2 gases), because the standard enthalpy change (AH) is positive for this reaction, meaning the products are more stable.

2) When we increase the volume of the container, the pressure decreases according to Boyle's law (P=k/V).

3) A decrease in pressure favors the side with the greater number of moles of gases. In this case, the product side has 2 moles of gas (H2 + O2), so the equilibrium will shift to the right.

4) Therefore, when the volume increases, the equilibrium will shift further in favor of the products (H2 and O2 gases).

The answer is a: It will shift in favor of the products.

Let me know if this makes sense! I can re-explain anything that is unclear.

Answer:

1. Gamma rays - f. Studying stars

2. Visible light - e. Light bulbs

3. Radio waves - b. Broadcasting radio stations

4. Infrared rays - d. Remote controls and night vision

5. X-Rays - a. Medical imaging

6. Ultraviolet rays - g. Tanning beds

7. Microwaves - c. Cell phones and radar