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3 years ago

How many mole of calcium (Ca) are in 9.00 x 1016 atoms of Ca?

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

5 0

Answer:

Number mole of calcium (Ca) = 1.4942 x 10⁻⁷ (Approx.)

Explanation:

Given:

Number of atom in (Ca) = 9 x 10¹⁶

Find:

Number mole of calcium (Ca)

Computation:

Number mole of calcium (Ca) = Number of atom in (Ca) / Avogadro's number

Number mole of calcium (Ca) = (9 x 10¹⁶) / (6.023 x 10²³)

Number mole of calcium (Ca) = 1.4942 x 10⁻⁷ (Approx.)

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How to balance this equation KClO4 → KCl + ?O2(g) and what type of reaction occurs?

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An element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. the heavier two isotopes have an abu

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3 years ago

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

pav-90 [236]

Answer: The initial concentration of hydrogen peroxide at the given temperature is 0.399 M

Explanation:

Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $2.54\times 10^{-4}s^{-1}$

t = time taken for decay process = 855 s

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.321 M

Putting values in above equation, we get:

$2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}$

$[A_o]=0.399M$

Hence, the initial concentration of hydrogen peroxide at the given temperature is 0.399 M

7 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

4 0

3 years ago