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2 years ago

How many grams are in 3.4 moles of Lead (IV) Sulfate (Pb(SO4)2)? ) Molar mass: 303.26 g/mol 131.0840

Chemistry

2 answers:

andrezito [222]2 years ago

6 0

Answer:

1031.08

Explanation:

You start with 3.4 moles and you know that 6.022 x 10 ^ 23 = 1 mole( Avogadro's Number), so you just multiply those 2 numbers and the moles will cancel leaving you with grams of Lead Sulfate.

Murrr4er [49]2 years ago

4 0

Answer:

1031.084g

Explanation:

n=m/M

m=n(M)

m = (3.4mol)(303.26g/mol)

1031.084g

Hope that helps

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73 g / 40.08 g = 1.8 mols of Ca in 73 grams

1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca

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3 years ago

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3 years ago

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The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of CaCO3 reacts with 0.4 moles * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2 produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 ---> Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol = 0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles reacts with 0.25 moles * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 = 0.04 moles of CO2 * 74 g/mol = 2.96 g

Percent yield = 2.5 g/ 2.96 g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

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2 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

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

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

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

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

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3 years ago

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Answer:

Increases in the frequency of intense rainfall,

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Explanation:

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3 years ago

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How many grams are in 3.4 moles of Lead (IV) Sulfate (Pb(SO4)2)? ) Molar mass: 303.26 g/mol 131.0840​

How many grams are in 3.4 moles of Lead (IV) Sulfate (Pb(SO4)2)? ) Molar mass: 303.26 g/mol 131.0840