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3 years ago

What is the volume of 1.60 grams of O2 gas at STP?

Mathematics

2 answers:

Mariulka [41]3 years ago

5 0

The answer is 1.12 or d

Verizon [17]3 years ago

3 0

<u>Answer:</u> The volume of given amount $O_2$ at STP is 1.12 L

<u>Explanation:</u>

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

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

Given mass of oxygen = 1.6 g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen}=\frac{1.6g}{32g/mol}=0.05mol$

At STP:

1 mole of a gas occupies 22.4 liters of volume.

So, 0.05 moles of oxygen gas will occupy = $\frac{22.4L}{1mol}\times 0.05mol=1.12L$ of volume.

Thus, the volume of given amount $O_2$ at STP is 1.12 L

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Answer:

Pedro spent 9 weeks with his uncle and his friend.

Step-by-step explanation:

First, you need to find the amount of weeks Pedro spent with his friend. The statement indicates that he spent two weeks more with his friend than with his uncle which means that you have to add the number of weeks he spent with his uncle plus 2, which you can show in a number line:

$3\frac{1}{2}$ 2

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$3\frac{1}{2}+2=5\frac{1}{2}$

Now, you can find the and answer by adding up the number of weeks he spent with his uncle plus the weeks he spent with his friend:

$3\frac{1}{2}$ $5\frac{1}{2}$

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$3\frac{1}{2} + 5\frac{1}{2}=9$

According to this, the answer is that Pedro spent 9 weeks with his uncle and his friend.

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3 years ago

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2 years ago

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}

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Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

$\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}$

Rationalize the denominator by multiplying both numerator and denominator by √10:

$\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}$

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

$\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}$

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