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3 years ago

What is the volume of 8.80 g of CH4 gas at STP?

Chemistry

1 answer:

Ann [662]3 years ago

6 0

Answer:

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.

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a couple of pennies have a mas of 5.2 grams. Asuming that the pennies are 100% percent copper,how many moles of copper are the p

dangina [55]

Answer:

5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)

Explanation:

mole conversions:

grams to moles => divide by formula wt.

moles to grams => multiply by formula wt.

gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)

This problem:

mass to moles => divide by formula wt.

mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)

5 0

2 years ago

If the H3O is 4.950 x 10-12 what is the ph?

romanna [79]

Answer:

pH = 11.3

Explanation:

From the question given above, the following data were obtained:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

The pH of a solution is defined by the following equation:

pH = –Log [H₃O⁺]

Thus, with the above formula, we can obtain the pH of the solution as follow:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

pH = –Log [H₃O⁺]

pH = –Log 4.950×10¯¹²

pH = 11.3

8 0

3 years ago

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.