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bogdanovich [222]
4 years ago
5

Iron has a density of 7.9 space straight g divided by cm cubed. How much mass is in a 10 space cm cubed sample of iron? Question

10 options: 17.9 g 79 g 1.26 g 0.79 g
Chemistry
1 answer:
dezoksy [38]4 years ago
6 0

Answer:

79 g

Explanation:

Density (d) of iron = 7.9 g/cm³

Volume of the iron (V) = 10 cm³

Mass of the iron (m) = ?

Solution:

Formula will be used

                            d = m/V

As we have to find mass so rearrange the above equation

                              m = dV . . . . . . . . . . . (1)

Put values in above equation 1

                         m = 7.9 g/cm³ x 10cm³

                         m = 79 g

So,

mass of iron = 79 g

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What is the fundamental unit of surface water hydrology?
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In the reaction of aluminum metal and oxygen to make aluminum oxide, how many grams of oxygen gas will react with 2.2 moles alum
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Answer:

52.8 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4Al + 3O2 —> 2Al2O3

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3

Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2.

Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.

Thus, 1.65 moles of O2 is needed for the reaction.

Finally, we shall determine the mass of O2 needed as shown below:

Mole of O2 = 1.65 moles

Molar mass of O2 = 2 × 16= 32 g/mol

Mass of O2 =?

Mole = mass/Molar mass

1.65 = mass of O2 /32

Cross multiply

Mass of O2 = 1.65 × 32

Mass of O2 = 52.8 g

Therefore, 52.8 g of O2 is needed for the reaction.

4 0
3 years ago
Determine the empirical formulas for compounds with the following percent compositions:
Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

3 0
3 years ago
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