Calculate the volume of a 0.0450 M HBr solution that is required to neutralize 120. mL of 0.0200 M Mg(OH)2. 2HBr Mg(OH)2 equation MgBr2 2H2O

To make 1 liter of a 10% NaCl solution from a solid stock, you will require the following materials and containers.MaterialsSolid NaClDistilled water1-Liter volumetric flask250-mL volumetric flask 2-beakersProcedureTo prepare 1 liter of a 10% NaCl solution, the following procedure should be followed:Measure out 100g of NaCl using a balance.

Measure the weight of an empty 250-mL volumetric flask.Add the NaCl to a 250-mL beaker and add a small amount of distilled water to it to dissolve the NaCl.Carefully pour the dissolved NaCl solution into the 250-mL volumetric flask. Add distilled water to the mark on the flask to make up the volume. Stopper the flask and invert it several times to mix the solution.Measure the weight of the 1-Liter volumetric flask.Add the 250-mL volumetric flask solution to a 1-Liter volumetric flask.Add distilled water to the mark on the flask to make up the volume.

Stopper the flask and invert it several times to mix the solution.The final volume of the solution will be 1 liter of a 10% NaCl solution.PrecautionsEnsure the NaCl has completely dissolved before adding more water to avoid making a less concentrated solution.Measure the weight of the volumetric flask before and after adding the solution to calculate the volume of solution that was added.Use distilled water to prepare the solution.

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Answer:

Answer:- Density of the substance is 0.7

i believe

Answer:

Phosphorus

Explanation:

Melting point is the exact temperature where it changes from solid to liquid. So if the temp 50°C, it is higher than 44°C meaning that the element already melted and is staying as a liquid.

Boiling point is the exact temperature where it changes from liquid to gas. So if the temp 300°C, it is higher than 28°C meaning that the element already evaporated and is staying as a gas.

Answer:

500ml

Explanation:

because 100 ×2 is 200 and the 300

500ml

The initial temperature of the gas was approximately 35.57 °C. find T1, we can rearrange the equation: T1 = (2.70 L * 23.00 °C) / 1.75 L.
Evaluating the expression, we get: T1 ≈ 35.57 °C.

To find the initial temperature of the gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Given that the gas is in a sealed container and the pressure is constant, we can rewrite the equation as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. Plugging in the values, we have: 2.70 L / T1 = 1.75 L / 23.00 °C.

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Answer:

5

Explanation:

Temperature calculates heat/the temperature in the specific location.

Answer:

A

Explanation:

You can see the top part looks kinda like methane, and the bottom part looks like hexane.

Answer:

A is correctvia A P E x

Explanation:

43 litres volume of hydrogen gas (H₂) are formed from the complete reaction of 17.7 g of C(s) .

In the given question we have a reaction of carbon solid with water. The reaction occuring carbon soild when it combines with water

basically form, water vapour. because here the state of water is gases so, it produce carbon monoxide gas with hydrogen gas.

C(s) + H₂O(g) -------> CO(g) + H₂(g)

Also, given the pressure at which hydrogen gas collected = 1.0 atm and

temperature is 355 Kelvin . We shall find out the value of volume of hydrogen gas.

mass of carbon soild = 17.7 g

molar mass of carbon = 12 g

now, we can find the moles of hydrogen gas in given reaction .

17.7 g ×( 1 mole of C(s) / 12.0 g of C(s) ) ( 1 mole of hydrogen / 1 mole of C(s) )

= 1.475 mol of hydrogen

Use the ideal gas equation, PV = n RT -----(1)

where, P --> pressure of hydrogen gas

V---> volume of hydrogen gas

T --> temperature of hydrogen gas

n --> numbers of mole of hydrogen

R ---> gas constant

and R = 0.0821 L atm /mol K

put all the known values in (1) we get ,

1 atm×V = 1.475 mole × 0.0821 L atm/mol K×355 K

V = 42.9896 L ~ 43 L.

Hence, the volume of hydrogen gas is 43 litres .

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The final temperature of the gas is approximately 40.96 Kelvin.

To determine the final temperature of the gas, we can use the ideal gas law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the given temperatures to Kelvin. We have:

Initial temperature: -125 degrees Celsius = 148 K (approximate)

Final temperature: Unknown

The initial conditions of the gas are as follows:

Initial pressure (P1) = 1.25 atm

Initial volume (V1) = 1250 mL = 1.25 L (since 1 L = 1000 mL)

Initial temperature (T1) = 148 K

The final conditions of the gas are as follows:

Final pressure (P2) = 50.0 atm

Final volume (V2) = 325 mL = 0.325 L

Final temperature (T2) = Unknown

Using the ideal gas law, we can set up the following equation:

(P1 * V1) / T1 = (P2 * V2) / T2

Substituting the known values:

(1.25 atm * 1.25 L) / 148 K = (50.0 atm * 0.325 L) / T2

Simplifying the equation:

T2 = (50.0 atm * 0.325 L * 148 K) / (1.25 atm * 1.25 L)

T2 = 40.96 K

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Answer:

A. a wild dog that finds and mates with a domestic dog.

Explanation:

Answer:

Carbon dioxide is a colorless and non-flammable gas at normal temperature and pressure. Although much less abundant than nitrogen and oxygen in Earth's atmosphere, carbon dioxide is an important constituent of our planet's air. A molecule of carbon dioxide (CO2) is made up of one carbon atom and two oxygen atoms.

Answer:

See Explanation

Explanation:

as potassium chloride(KCI) dissolves in water, the potassium and chloride  ions are formed and hydrated. Ion-dipole forces attract the positive(Hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative(oxygen) ends to the positive potassium ions

Ten transition metals along with their respective atomic masses:

1. Scandium (Sc) - Atomic Mass: 44.9559 u

2. Titanium (Ti) - Atomic Mass: 47.867 u

3. Vanadium (V) - Atomic Mass: 50.9415 u

4. Chromium (Cr) - Atomic Mass: 51.9961 u

5. Manganese (Mn) - Atomic Mass: 54.9380 u

6. Iron (Fe) - Atomic Mass: 55.845 u

7. Cobalt (Co) - Atomic Mass: 58.9332 u

8. Nickel (Ni) - Atomic Mass: 58.6934 u

9. Copper (Cu) - Atomic Mass: 63.546 u

10. Zinc (Zn) - Atomic Mass: 65.38 u

Transition metals are elements found in the d-block of the periodic table. They exhibit characteristic properties such as the formation of colored compounds, variable oxidation states, and the ability to form complex ions.

These metals often have multiple stable oxidation states due to the availability of different energy levels within their d-orbitals. This property allows them to form a wide range of compounds and exhibit various chemical behaviors.

Transition metals also have high melting and boiling points, good thermal and electrical conductivity, and are often malleable and ductile.

Their atomic masses vary depending on the specific isotope of the element, which may have a different number of neutrons. The atomic mass represents the average mass of all naturally occurring isotopes of an element, taking into account their relative abundance.

These transition metals play vital roles in various industrial processes, including catalysis, electronics, alloys, and the production of pigments and dyes. Their unique properties and versatility make them important elements in many fields of science and technology.

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Answer:

It is 'A' i.e., Sodium magnesium, aluminium

Elements tend to lose and gain electrons to achieve electronic stability. Lithium, sodium, and potassium are the elements that form +1 ions. Thus, option C is correct.

Cations are said to be the ions formed by donating their outermost valence electron that in return results in a positive charge on the element. Cations are said to be formed when the electron is more in number in the valence shell.

The electronic configuration is given as:

Here, there is one electron extra in the outer shell of the atom that can be donated to other atoms that are called anions (electron acceptors). Li, Na, and K can donate one electron each and hence will acquire a +1 charge.

Therefore, option C. lithium, sodium, and potassium form a +1 charge.

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Answer:

21 March

Explanation:

hope its helpful

Answer:

50°F,  70°F

Explanation:

Answer:

50 and 70 degrees F

Explanation:

got it right on the quiz on edge

Answer: F

Explanation:

we can prevent soil erosion by reducing the amount of water and blockages that might build water up and erode the soil around farmland/other

Answer:

for pH 13 it = strong based the ph 2 is= weak acid

Explanation:

CBr₄ has no net dipole moment.

Net dipole moment is the total sum of the bond moments in a molecule. It is a measure of the polarity of a molecule. In other words, the dipole moment of a molecule indicates the direction and magnitude of its polarity. A molecule has no net dipole moment when it is nonpolar.

This happens when the dipole moments of its polar bonds cancel each other out. There are various types of molecules, including linear, bent, trigonal planar, tetrahedral, and pyramidal. These molecules can be polar or nonpolar, depending on their geometrical structure and bond polarity.

CBr₄ has a tetrahedral shape, with four Bromine atoms and a Carbon atom at the vertices of the tetrahedron. The C-Br bonds are polar because Bromine is more electronegative than Carbon. However, the four C-Br bond dipoles are arranged in a tetrahedral structure, such that they cancel out each other. Thus, CBr₄ has no net dipole moment.

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: PM2.5 is defined as the mass concentration of particles in the air less than or equal to 2.5 micrometers in diameter.Question 15: False, carbon dioxide (CO2) is not considered a criteria air pollutant.

Question 16: The closest answer is 50%, but the exact percentage is not provided in the question.Question 17: The term "photochemical smog" is most synonymous with ozone (O3), which is a criteria air pollutant.Question 18: True, attainment of ambient air quality standards requires that measured concentrations at all monitoring stations within an air district are below ambient air standards.

Question 14 asks about the definition of PM2.5. PM2.5 refers to particulate matter with a diameter less than or equal to 2.5 micrometers. It represents the mass concentration of particles suspended in the air, which are small enough to be inhaled into the respiratory system and can have adverse health effects.

Question 15 states whether carbon dioxide (CO2) is a criteria air pollutant. Criteria air pollutants are a set of pollutants regulated by environmental agencies due to their detrimental impact on air quality and human health. However, carbon dioxide is not considered a criteria air pollutant because it does not directly cause harm to human health or the environment in the same way as pollutants like ozone or particulate matter.

Question 16 asks about the percentage of carbon monoxide (CO) emissions from mobile sources in Santa Clara County. While the exact percentage is not provided in the question, the closest answer option is 50%. However, it is important to note that the precise percentage may vary depending on specific local conditions and emissions sources.

Question 17 inquires about the criteria air pollutant most synonymous with the term "photochemical smog." Photochemical smog is primarily associated with high levels of ground-level ozone (O3). Ozone is formed when nitrogen oxides (NOx) and volatile organic compounds (VOCs) react in the presence of sunlight, creating a hazy and polluted atmospheric condition.

Question 18 addresses the concept of "attainment" of ambient air quality standards. To achieve attainment, measured concentrations of pollutants at all monitoring stations within an air district must be below the established ambient air quality standards. This ensures that the air quality in the given area meets the required standards for protecting human health and the environment.

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The speed in number of meters per minute the spacecraft travel at the speed of 5.2 km per second is 3,12,000 meters per minute.

The speed is 5.2 km per second

We know that 1 kilometer per second is equal to 60000 meter per minute

So converting, we get 5.2×60000= 3,12,000 meters per minute

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0.10 M will be the concentration of A after 80 min.

We need to use the equation for first order reactions, which is: ln[A]t = -kt + ln[A]0, where [A]t is the concentration of A at time t, k is the rate constant, and [A]0 is the initial concentration of A.
We are given that the half-life of the reaction is 20 minutes, which means that k = ln2/20 = 0.03465 min^-1.
We can now use this value of k to find the concentration of A after 80 minutes:
ln[A]80 = -0.03465 x 80 + ln(1.6)
ln[A]80 = -2.772 + 0.470
ln[A]80 = -2.302
To get the concentration of A, we need to take the antilog of this value:
[A]80 = e^-2.302
[A]80 = 0.099 M
Therefore, the answer is (C) 0.10 M.
Substance A undergoes a first-order reaction A → B with a half-life of 20 minutes at 25 °C. The initial concentration of A is 1.6 M. To determine the concentration of A after 80 minutes, we can use the half-life concept. Since 80 minutes is equivalent to 4 half-lives (80 minutes / 20 minutes per half-life), we can calculate the concentration as follows:
1st half-life (20 min): 1.6 M / 2 = 0.8 M
2nd half-life (40 min): 0.8 M / 2 = 0.4 M
3rd half-life (60 min): 0.4 M / 2 = 0.2 M
4th half-life (80 min): 0.2 M / 2 = 0.1 M
Therefore, the concentration of A after 80 minutes will be 0.1 M (Option C).

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Answer:

The answer is C

Explanation:

I did it on edge

Answer:

OB

Explanation:

interview with many smokers ,in this way she would directly be able to tell the effects

The volume of 10.0 M nitric acid required to prepare 179.0 mL of 0.37 M nitric acid solution is 6.52 mL (rounded to two decimal places).  

The volume of 10.0 M nitric acid required to prepare 179.0 mL of 0.37 M nitric acid solution is 6.52 mL.What is Molarity?Molarity, also known as concentration, is the number of moles of solute in one liter of solution. It is measured in mol/L (molarity) and is represented as M.

How to calculate Molarity?

Molarity can be calculated using the following formula: M = n / V Where, M is the molarity n is the number of moles of solute V is the volume of the solution in liters Given, Molarity of nitric acid = 10.0 M Volume of the solution = 179.0 mL = 0.179 L Required Molarity of nitric acid = 0.37 M We can use the following formula to find the volume of the 10.0 M nitric acid:

M1V1 = M2V2

Where,M1 is the initial molarity of the solutionV1 is the initial volume of the solutionM2 is the final molarity of the solutionV2 is the final volume of the solution Let us substitute the values in the formula:M1V1 = M2V210.0 M × V1 = 0.37 M × 0.179 LV1 = (0.37 M × 0.179 L) / 10.0 MV1 = 0.00661 L = 6.61 mL

Therefore, the volume of 10.0 Mole nitric acid required to prepare 179.0 mL of 0.37 M nitric acid solution is 6.52 mL (rounded to two decimal places).

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The diamagnetic materials, ferromagnetic materials, and paramagnetic materials are the three categories that classify the magnetic properties of matter.

Magnetic properties of matter can be grouped into three distinct categories: diamagnetic, ferromagnetic, and paramagnetic materials. Diamagnetic materials exhibit weak or no magnetic response when exposed to a magnetic field, causing them to be repelled by the field.

On the other hand, ferromagnetic materials display strong magnetic behavior, becoming permanently magnetized in the presence of a magnetic field. These materials retain their magnetism even after the field is removed. Paramagnetic materials fall in between, showing a temporary attraction to the magnetic field but not becoming permanently magnetized.

These materials exhibit a weak magnetic response and lose their magnetism once the external magnetic field is removed. Understanding these classifications is crucial for various applications in physics, materials science, and engineering.

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The vertical jumps in the plot at the melting and boiling points represent the abrupt increase in entropy that occurs during phase transitions from solid to liquid and from liquid to gas.

The plot shows vertical jumps at the melting and boiling points because the transition from one phase to another involves a change in the entropy (S) of the substance. At the melting point, the substance undergoes a phase transition from solid to liquid, resulting in an abrupt increase in entropy. Similarly, at the boiling point, the substance transitions from liquid to gas, leading to another sudden increase in entropy.

The melting and boiling points represent phase transitions where the substance undergoes a change in its physical state. During these transitions, the arrangement and movement of particles change, leading to a significant alteration in entropy.

At the melting point, a solid substance absorbs energy to overcome intermolecular forces and transition into a liquid state. This transition involves an increase in entropy as the particles gain more freedom to move around, leading to a jump in the entropy value on the plot.

Similarly, at the boiling point, a liquid substance absorbs energy to overcome intermolecular forces and transition into a gaseous state. This transition also results in an increase in entropy as the particles gain even more freedom of movement and are dispersed in the gas phase, causing another vertical jump in the entropy value on the plot.

In summary, the vertical jumps in the plot at the melting and boiling points represent the abrupt increase in entropy that occurs during phase transitions from solid to liquid and from liquid to gas.


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