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yulyashka [42]
3 years ago
15

What is the volume, in liters, of 0.500 mol of c3h8 gas at stp? (hint..use avogadro’s principle to solve this)?

Chemistry
2 answers:
laila [671]3 years ago
8 0

11.2.............................................................

wolverine [178]3 years ago
3 0
When we are at STP conditions, we can use this conversion: 1 mol= 22.4 L

0.500 mol C₃H₈ (22.4 L/ 1 mol)= 11.2 L
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The balanced chemical reaction is:

<span>2Na + 2H2O → 2NaOH + H2
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We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.

53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na

1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>
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For the following problems, identify the substance as either an element, a compound, or a mixture. Write E for Element, C for co
andre [41]

Answer:

Explanation:

34. E

35. M

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39. C

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3 years ago
How important is it for scientists and engineers to be able to control the design of
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3 years ago
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The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a
Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g

Carbon-14 is 1.6\times 10^{-10}\ \% of the carbon in the body. So,

\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g

Also,

14 g of Carbon-14 contains  6.023\times 10^{23} atoms of carbon-14

So,  

2.48\times 10^{-8}\ g of Carbon-14 contains  \frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8} atoms of carbon-14

Atoms of carbon-14 =  1.07\times 10^{15}

Given that:

Half life = 5730 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac{ln\ 2}{5730}\ years^{-1}

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = 3.154\times 10^7 s

So, The rate constant, k = \frac{0.00012}{3.154\times 10^7} s⁻¹ = 3.8\times 10^{-12}\ s^{-1}

Thus, decay events per second = K\times atoms decayed = 3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s = 4066 decay/s

4 0
3 years ago
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