How many grams of magnesium bromide (MgBry) are in 5.38 x 1024 formula units
of the substance?

The statements that are applicable to Hess's Law include:

Hess's Law is also referred to as Hess's law of constant heat summation and it was named after a Swiss-born Russian chemist called Germain Hess.

Hess's Law states that the energy change experienced in an overall chemical reaction is equal to the sum of the energy changes in each chemical reactions that it contains.

In conclusion, the enthalpy change of a chemical reaction which is the heat of reaction at constant pressure is independent of the pathway between its initial and final states.

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Answer:

The Changes I see is the Hot glass have less air bubbles than the Cold Glass. The Cold Glass has more air bubbles than the Hot Glass.

I hope this helps. I tried my best!! Good Luck!!

The heat of reaction is the quantity of heat absorbed or emitted in a reaction.

The heat of reaction is the quantity of heat absorbed or emitted in a reaction. This question is incomplete hence the numerical value of the heat of reaction can not be determined.

However, the heat of reaction is determined from the mass of the reactants, the temperature change in the calorimeter and the number of moles of the limiting reactant.

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Answer:  For the reaction 2HCl(g) ==> H2(g) + Cl2(g), the Keq = [H2(g)][Cl2(g)]/[HCl(g)]^2

Explanation:

Answer:

b option sir sidd dichhi good book goo

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore,  which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

[tex]\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\[/tex]

Where the short tons are cancelled out as well as the grams, in order to obtain:

[tex]0.103 otz[/tex]

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Answer:

1 mole contains 6.02 ×10^23 atoms

1 mole of Sulphur will contain 6.02 ×10^23 atoms

2moles of. Oxygen will contain 2 x 6.02×10^23 atoms. This gives 12.04 × 10^23 atoms

Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O

Explanation:

Answer: Celsius, Fahrenheit, Kelvin, and Rankine.

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750 x 14,0067 u = 10,505.025 g

Answer:

1.8x10_24 moleclues

Explanation:

because you have to divide the numbers  and then times them together

what's up? i don't think D words it that well, but i would think it's D since none of the other answers make sense

Answer:

so you can give it

Explanation:

Answer:

3 significant figures

Explanation:

Thus 2° C, converts to 275 K, which has 3 significant figures. Rounding Error When you do a long series of calculations, try to do them all in one set of operations.

Answer:

the sum of average kinetic molecules of the body is called temperature..

Answer:

it's a I took the test

Explanation

Answer:

A) The new substance will need more energy to form its chemical bonds than the old substance will release.

Explanation:

Answer:

sorry man i cant help you i tried my best im so so so sorry

Explanation:

[tex]\tt \rho=\dfrac{m}{v}=\dfrac{7\times 10^2~g}{1.3~g/L}=538~L[/tex]

From what we know, we can confirm that the difference measured by the experiment in the heat released when compared to the hydrogenation of an imaginary cyclohexatriene is called the stabilization energy of benzene.

This is a term used to describe the difference in measured energy of a chemical structure compared to the theoretical energy amounts expected to be measured. In this experiment, the heat that was released did not match the calculated energy expected, which is a prime example of stabilization energy at work.

Therefore, we can confirm that the difference measured by the experiment is called the stabilization energy of benzene.

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Based on the mass ratio of reactants, 23.5 g of sodium nitrate is produced from the reactants.

The mass ratio of a reaction is the ratio of the masses of reactants wilhich combine to form products.

The mass of sodium nitrate is determined from the mass ratio.

The equation of the reaction gives the mass ratio as follows:

NaCl + AgNO3 ----> AgCl + NaNO3

Molar mass of NaCl = 58.5 g

Molar mass of AgNO3 = 170 g

Molar mass of AgCl = 143.5

Molar mass of NaNO3 = 85 g

There are 98 grams of sodium chloride, 47 grams of silver nitrate, therefore silver nitrate is the limiting reactant.

47 grams of silver nitrate will produce 85/170 × 47 = 23.5 g of sodium nitrate.

Therefore, 23.5 g of sodium nitrate is produced from the reactants.

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Answer:

salinity is the concentrations of salts in water or soils, chlorinity is a measure of the concentration of halides in one kilogram of seawater. Explanation:
srry if its wrong ;-;

Answer:

Chlorinity and salinity are both measures of the saltiness of sea water. The relationship can be expressed mathematically, as salinity is equivalent to 1.80655 times the chlorinity.

Conclusion:

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Thank You!

Answered by: ms115

The ΔG for the reaction shown is -11.76 kJ/mol.

The equilibrium constant is a value that shows the extent to which reactants are converted to products in a reaction. We have to first obtain the equilibrium constant as follows;

K = (0.750)^2/(0.160) (0.100) = 35

Now;

Using the formula;

ΔG = -RTlnK

ΔG = -(8.314 × 398 × ln 35)

ΔG = -11.76 kJ/mol

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Answer:

58.92 g EDTA

Explanation:

315.1 mL = .3151 L    

M = Moles / Liter

.3151 L  x 0.5 mol EDTA  x  374 g EDTA  =  58.92 g EDTA

                1 L EDTA             1 mol EDTA

The amount in grams of EDTA needed to make 315.1 mL of a 0.05 M is equal to 5.89 g.

In a 0.05M solution, there are 0.05 moles of the substance per 1 liter, that is, in 315.1 ml we have:

                                        [tex]\frac{0.05 mol}{xmol}= \frac{1000ml}{315.1ml}[/tex]

                                           [tex]x = 0.016 mol[/tex]

Thus, 0.016 mol EDTA of  is needed to prepare the solution.

To calculate the mass in an amount of moles, we use the molar mass, so that:

                                          [tex]m = MM \times mol[/tex]

                                         [tex]m = 374 \times 0.016[/tex]

                                             [tex]m = 5.89 g[/tex]

So, the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M is equal to 5.89 g.

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In 145.45 grams of water there are 160mg of 0.0011% Pb.

To resolve this issue, it is necessary to use the concentration of solutions, which indicates the percentage amount of one solvent or solute in relation to the other.

In the case of the question, we have that Pb is at a concentration of 0.0011%, that is, for every 100 grams of water we have 0.0011 grams of Pb.

So, to calculate the amount of water that we have 160mg of Pb, we will make the following expression:

                                           Pb           H2O

                                            [tex]\frac{0.0011g}{160\times 10^ {-3}g} =\frac{1g}{x}[/tex]

                                            [tex]x = 145.45g[/tex]

So, in 145.45 grams of water there are 160mg of 0.0011% Pb.

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Answer:

where are the following organisms

The balanced equation is:

4 NO(g) + O₂(g) ⇒ 2 N₂O₃(g)

The expression of the rate in terms of change in concentrations are:

[tex]v = \frac{-\Delta [NO]}{4\Delta t } = \frac{-\Delta [O_2]}{1\Delta t } = \frac{\Delta [N_2O_3]}{2\Delta t }[/tex]

When N₂O₃ is forming at 0.221 M/s, NO is disappearing at 0.442 M/s.

The reaction rate is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.

4 NO(g) + O₂(g) ⇒ 2 N₂O₃(g)

We will need to consider the stoichiometric coefficient of each species.

[tex]v = \frac{-\Delta [NO]}{4\Delta t } = \frac{-\Delta [O_2]}{1\Delta t } = \frac{\Delta [N_2O_3]}{2\Delta t }[/tex]

The molar ratio of NO to N₂O₃ is 4:2.

[tex]\frac{0.221molN_2O3}{L.s} \times \frac{4molNO}{2molN_2O_3} = \frac{0.442 molNO}{L.s}[/tex]

The balanced equation is:

4 NO(g) + O₂(g) ⇒ 2 N₂O₃(g)

The expression of the rate in terms of change in concentrations are:

[tex]v = \frac{-\Delta [NO]}{4\Delta t } = \frac{-\Delta [O_2]}{1\Delta t } = \frac{\Delta [N_2O_3]}{2\Delta t }[/tex]

When N₂O₃ is forming at 0.221 M/s, NO is disappearing at 0.442 M/s.

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