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Katyanochek1 [597]
2 years ago
11

What is the yield of uranium from 2.50 kg U3O8?

Chemistry
1 answer:
mr_godi [17]2 years ago
8 0

Answer: Thus the yield of uranium from 2.50 kg U_3O_8 is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number (6.023\times 10^{23}) of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}

moles of U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol    (1kg=1000g)

As 1 mole of U_3O_8 contains = 3 moles of U

2.97 mole of U_3O_8 contains = \frac{3}{1}\times 2.97=8.91moles moles of U

Mass of Uranium=moles\times {\text {Molar mass}}=8.91mol\times 238g/mol=2120g=2.12kg  

 ( 1kg=1000g)

Thus the yield of uranium from 2.50 kg U_3O_8 is 2.12 kg

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2 years ago
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Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4
aleksandr82 [10.1K]

Answer:

Molarity for the sulfuric acid is 0.622 M

Explanation:

When we neutralize an acid with a base, molarity of both . both volume are the same. The formula is:

M acid . volume of acid = M base . volume of base

M acid = unknown

Volume of acid = 17 mL

Volume of base = 45 mL

M base = 0.235 M

Therefore, we replace:  M acid . 17 mL = 0.235 M . 45 mL

M acid = (0.235 M . 45 mL) / 17 mL

M acid = 0.622 M

6 0
2 years ago
Determine the number of moles in 4.21 x 10^23 molecules of CaCl2
Paha777 [63]
<h3>Answer:</h3>

0.699 mole CaCl₂

<h3>Explanation:</h3>

To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

But, 1 mole of a compound contains  6.022 x 10^23 molecules

In this case;

we are given 4.21 × 10^23 molecules of CaCl₂

Therefore, to get the number of moles

Moles = Number of molecules ÷ Avogadro's constant

          = 4.21 × 10^23 molecules ÷  6.022 x 10^23 molecules/mole

          = 0.699 mole CaCl₂

Hence, the number of moles is 0.699 mole of CaCl₂

7 0
2 years ago
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0
2 years ago
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Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?
balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

<h3>C 8.09 SO2 gas </h3>

8 0
2 years ago
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