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fredd [130]
4 years ago
5

Determine the volume of a 75-gram sample of gold at stp

Chemistry
1 answer:
Anuta_ua [19.1K]4 years ago
7 0
Let V = the volume of the sample at STP.

The mass is  m = 75 g
The density of gold is 19.32 g/cm³

Because density = mass/volume, therefore
(19.32 g/cm³) = (75 g)/(V cm³)
V = (75 g)/(19.32 g/cm³) = 3.882 cm³

Answer: 3.882 cm³
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Calculate the molarity of a MgSO4, solution
Gala2k [10]

Answer:

0.058M is molarity of the solution

Explanation:

Molarity is defined as the ratio between the moles of solute (In this case, MgSO₄) and the liters of solution.

In the problem:

<em>Moles MgSO₄: 0.32 moles</em>

<em>Liters solution: 5.5L</em>

<em />

Molarity is:

0.32 moles / 5.5L =

0.058M is molarity of the solution

6 0
3 years ago
How much 0.02 m kmno4 solution should be needed if the solutions tested have a composition of 3% h2o2 by mass?
Nostrana [21]
When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:

0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4

Solving for V,
V = 30.69 L
6 0
3 years ago
The density of an unknown gas at 98°C and 740 mmHg is 2.50 g/L. What is the molar mass of the gas with work showed?
noname [10]

Answer:

78.2 g/mol  

Step-by-step explanation:

We can use the <em>Ideal Gas Law</em> to solve this problem:

       pV = nRT

Since n = m/M, the equation becomes

      pV = (m/M)RT     Multiply each side by M

   pVM = mRT               Divide each side by pV

        M = (mRT)/(pV)

Data:

ρ = 2.50 g/L

R = 0.082 16 L·atm·K⁻¹mol⁻¹

T =98 °C

p = 740 mmHg

Calculation:

(a)<em> Convert temperature to kelvins </em>

T = (98 + 273.15) = 371.15 K

(b) <em>Convert pressure to atmospheres </em>

p = 740 × 1/760 =0.9737 atm

(c) <em>Calculate the molar mass </em>

Assume V = 1 L.

   Then m = 2.50 g

            M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)

                = 76.14/0.9737

                = 78.2 g/mol

3 0
3 years ago
Plzzz help me! I will do anything ;) <br><br> How many molecules are there in 45g of caffeine
babymother [125]

Answer:

45 IS THE ANSWER

Explanation:

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6 0
3 years ago
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
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