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Hitman42 [59]
2 years ago
7

What is the maximum mass, in kilograms, of (NH4)2U2O7 that can be formed from the reaction of 100 kg of water and 100 kg of ammo

nia with 481 kg of UO2SO4?
Chemistry
1 answer:
V125BC [204]2 years ago
5 0

Mass of (NH₄)₂U₂O₇ : 410.05 kg

<h3>Further explanation</h3>

Reaction

2UO₂SO₄ + 6NH₃ + 3H₂O → (NH₄)₂U₂O₇ + 2(NH₄)₂SO₄

MW UO₂SO₄ :  366.091

MW (NH₄)₂U₂O₇ : 624.131

MW H₂O :  18.0153

MW NH₃ : 17.0306

mol of 100 kg water :

\tt \dfrac{100}{18.0153}=5.55

mol of 100 kg ammonia :

\tt \dfrac{100}{17.036}=5.87

mol of UO₂SO₄ :

\tt \dfrac{481}{366.091}=1.314

Limiting reactants : smallest mol ratio(mol : coefficient)

\tt \dfrac{5.55}{3}\div \dfrac{5.87}{6}\div \dfrac{1.314}{2}=1.85\div 0.98\div 0.657

UO₂SO₄ ⇒ Limiting reactants

mol (NH₄)₂U₂O₇ : mol UO₂SO₄

\tt \dfrac{1}{2}\times 1.314=0.657

mass (NH₄)₂U₂O₇

\tt 0.657\times 624.131=410.05

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A 35.0 mL sample of 1.00 M KBr and a 60.0 mL sample of 0.600 M KBr are mixed. The solution is then heated to evaporate water unt
Katarina [22]

Answer: The molarity of KBr in the final solution is 1.42M

Explanation:

We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.

We will first calculate the number of moles of KBr in the individual sample before mixing together

In the first sample:

Volume (V) = 35.0 mL

Concentration (C) = 1.00M

Number of moles (n) = C × V

n = (35.0mL × 1.00M)

n= 35.0mmol

For the second sample

V = 60.0 mL

C = 0.600 M

n = (60.0 mL × 0.600 M)

n = 36.0mmol

Therefore, we have (35.0 + 36.0)mmol in the final solution

Number of moles of KBr in final solution (n) = 71.0mmol

Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.

Therefore,

Final volume of solution (V) = 50mL

Number of moles of KBr in final solution (n) = 71.0mmol

From

C = n / V

C= 71.0mmol/50mL

C = 1.42M

Therefore, the molarity of KBr in the final solution is 1.42M

5 0
2 years ago
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
What causes the biggest change mass or speed?
Norma-Jean [14]

Answer:

Explanation if an object is in motion and more force is applied to it, the object will begin moving faster. If two objects have the same mass and a greater force is applied to one of the objects, the object which receives the greater force will change speeds more quickly.:

6 0
2 years ago
Calculate the mole fraction of Ba Cl 2 in an aqueous solution prepared by dissolving 0.400 moles of Ba Cl 2 in 850.0 g of water.
DaniilM [7]

Answer:

0.0084

Explanation:

The mole fraction of BaCl₂ (X) is calculated as follows:

X = moles BaCl₂/total moles of solution

Given:

moles of BaCl₂ = 0.400 moles

mass of water = 850.0 g

We have to convert the mass of water to moles, by using the molecular weight of water (Mw):

Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol

moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol

The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):

total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol

Finally, we calculate the mole fraction:

X = 0.400 mol/47.6 mol = 0.0084

4 0
2 years ago
In normal conditions, warm water "piles up" in the Western Pacific Ocean.<br><br> True<br> False
kakasveta [241]
In normal conditions, warm water does "pile up" in the" Western Pacific Ocean.
8 0
2 years ago
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