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3 years ago

Convert 23.064 mass to mg

Chemistry

1 answer:

Ierofanga [76]3 years ago

3 0

ANSWER: 23064 mg

EXPLANATION: To convert grams to milligrams, we multiply by 1000.

23.064 g x 1000 = 23064 mg

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Answer: The mass of lead iodide produced is 9.22 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

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3 years ago

An element has an atomic mass of 69.66 amu. It has two

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Answer: 63.26%

Explanation:

If we let the abundance of the first isotope be x, then:

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Which is equal to 63.26%

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(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n

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Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

$5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O$

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According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

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