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2 years ago

Metric dimensional analysis worksheet problem: 645 kg to g

Chemistry

1 answer:

Andrews [41]2 years ago

3 0

Answer:

645000 grams my dude

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Oxidation
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2 years ago

Find the molarity of the following solutions: 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution.

vredina [299]

M(molarity)=moles/volume
M=.5mol/.05L=10M

4 0

2 years ago

What is the density of an unknown object that has a volume of 4cm. and a mass of 10g.

Dmitry [639]

The density would be 2.5g/cm^3

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2 years ago

Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4

Anvisha [2.4K]

Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

Pb(SO₄)₂ : Li₂SO₄

1 : 2

2.25 : 2/1×2.25 = 4.5 mol

LiNO₃ : Li₂SO₄

4 : 2

9.62 : 2/4×9.62 = 4.81 mol

Pb(SO₄)₂ produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of Li₂SO₄.

7 0

2 years ago

The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con

bonufazy [111]

Answer: The concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

Explanation:

All the radioactive reactions follows first order kinetics.

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

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

We are given:

$t_{1/2}=3.82days$

Putting values in above equation, we get:

$k=\frac{0.693}{3.82}=0.181days^{-1}$

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 = $0.181days^{-1}$

t = time taken for decay process = 3.00 days

$[A_o]$ = initial amount of the reactant = $1.45\times 10^{-6}mol/L$

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}$

$[A]=3.83\times 10^{-30}mol/L$

Hence, the concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

7 0

3 years ago