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anyanavicka [17]
2 years ago
15

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

molarity of nh3 in the solution is ________ m.
Chemistry
2 answers:
iris [78.8K]2 years ago
8 0

To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ<span> H2O)                        ---> 1</span>

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

ρ NH3 = 0.680 g / mL

Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

Therefore the molarity of NH3 in the solution is:

<span>Molarity = 0.881 mol / [(22.07 mL  + 250 mL) * (1L / 1000 mL)</span>

<span>M = 3.238 mol/L = 3.24 M</span>

blsea [12.9K]2 years ago
7 0

<u>Answer:</u> The molarity of NH_3 in  the solution is 3.24 M.

<u>Explanation:</u>

To calculate the molarity of ammonia in solution, we first need to find the volume of solution and we will calculate the volume of solution by using the density of the solution.

To calculate the volume of an object, we use the formula:

Density=\frac{\text{Mass of an object}}{\text{Volume of an object}}

We are given:

Density = 0.974g/mL

Mass of the solution = (15 + 250)g = 265g

Putting values in above equation, we get:

0.974g/mL=\frac{265g}{Volume}

Volume of the solution = 272 mL

Now, to calculate the molarity of the solution, we use the formula:

Molarity=\frac{w\times 1000}{M\times V\text{(in mL)}}

where,

w = weight of ammonia = 15 g

M = molar mass of ammonia = 17g/mol

V = Volume of the solution = 272 mL

Putting values in above equation, we get:

\text{Molarity of ammonia}=\frac{15\times 1000}{17\times 272}\\\\\text{Molarity of ammonia in solution}=3.24M

Hence, molarity of ammonia in solution is 3.24M

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Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 78 mL of wa
Vaselesa [24]

Answer:

We need 1.1 grams of Mg

Explanation:

Step 1: Data given

Volume of water = 78 mL

Initial temperature = 29 °C

Final temperature = 78 °C

The standard heats of formation

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Step 2: The equation

The heat is produced by the following reaction:

Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)

Step 3: Calculate the mass of Mg needed

Using the standard heats of formation:

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)

−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg

(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required

(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg

We need 1.1 grams of Mg

7 0
2 years ago
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
Maslowich

Answer:

94.7 %

Explanation:

The balanced chemical reaction is as follows -

2 S + 3 O₂ -----> 2 SO₃

Hence ,

The number of moles according to the chemical reaction is -

Moles of S = 2

Moles of O₂ = 3

Since ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

For Sulfur ,

w = 6.0 g

as we know the molecular mass of S = 32 ,

Hence ,

moles are calculated by using the above formula , and putting the respective values ,

n = w / m  = 6 / 32 = 0.1871 mol

similarly for Oxygen ,

w = 5.0 g

as we know the molecular mass of O = 32 g/mol ,

Hence ,

moles are calculated by using the above formula , and putting the respective values ,

n = w / m  = 5 / 32 = 0.15625 mol

Comparing the moles with the values from the the equation ,

Since , S is in excess ,

Therefore , O₂ is the limiting reagent ,

Hence O₂ will determine the moles of the product ,

Therefore ,

Using unitary method ,

From the chemical reaction ,

3 mol of O₂ will give 2 mol of SO₃

and ,

1 mol of O₂ will give 2 / 3 mol of SO₃

Therefore ,

0.15625 mol O₂ will give 2 / 3 * 0.15625 mol of SO₃

Hence ,

Moles of  SO₃ produced = 0.1042 mol

since ,

n = w / m

w = n * m

molecular mass of SO₃ = 80 g/mol

w = Mass of SO₃ =  0.1042 mol * 80 g/mol = 8.340 g

The percentage yield is calculated as the actual yield divided by the theoretical yield multiplied by 100 .

Hence ,

Percentage yield = actual yield / theoretical yield * 100

From the question ,

actual yield = 7.9 g of  SO₃

As calculated above ,

theoretical yield = 8.340 g

Percentage yield =  7.9 g / 8.340 g * 100 = 94.7 %

6 0
2 years ago
At 1 atm, how many moles of co2 are released by raising the temperature of 1 liter of water from 20∘c to 25∘c? express your answ
tekilochka [14]
Answer is: 0,0030 mol of carbon dioxide.
Carbon dioxide solubility in water at 20°C and 1 atm is: 3,8·10⁻² mol/L.
Carbon dioxide solubility in water at 25°C and 1 atm is: 3,5·10⁻² mol/L.
Difference is: 0,038 mol/L - 0,035 mol/L = 0,003 mol/L.
V(carbon dioxide) = 1 L.
n(carbon dioxide) = V · c.
n(carbon dioxide) = 1 L · 0,003 mo/L.
n(carbon dioxide) = 0,0030 mol.
4 0
2 years ago
Read 2 more answers
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igor_vitrenko [27]
When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
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P = (2)(1 m)(8.314)(298 K)
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P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
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7 0
2 years ago
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ICE Princess25 [194]

Answer:

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2 years ago
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