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2 years ago

How do felsic igneous rocks

Chemistry

1 answer:

Svetradugi [14.3K]2 years ago

8 0

Answer:

A. Felsic igneous rocks are less dense than mafic igneous rock

Explanation:

"Felsic rocks are composed of larger quantities of silicates and therefore are less dense. Felsic magma is less dense and more viscous than mafic magma." - study.com

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Please help me with chemistry

Murljashka [212]

Answer:

60.052 g/mol (molar mass of vinegar)

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3 years ago

Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.50 L and the temperature is increased to 3

pickupchik [31]

First convert celcius to Kelvin.

20 + 273 = 293K

31 + 273 = 304K

Now we can set up an equation based on the information we have.

V1 = 5

P1 = 365

T1 = 293

V2 = 5

P1 = x

T2 = 304

The equation be: $\frac{(5)(365)}{293} = \frac{5x}{304}$

Now just solve.

1825/293 = 5x/304

Cross multiply.

554800 = 1465x

Divide both sides by 1465

x = 378.7030717 which can then be rounded to 378.7 mmHg

4 0

3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: 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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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4 years ago

For the following pairs of ions, use the concept that a chemical compound must have a net charge of zero to predict the formula

Dafna11 [192]

$\\ \sf\longmapsto FeP$

$\\ \sf\longmapsto Fe_2S_3$

$\\ \sf\longmapsto FeCl_2$

$\\ \sf\longmapsto MgCl_2$

$\\ \sf\longmapsto MgO$

$\\ \sf\longmapsto Mg_3N_2$

$\\ \sf\longmapsto Na_3P$

$\\ \sf\longmapsto Na_2S$

$\\ \sf\longmapsto CoO$

$\\ \sf\longmapsto Co_2S_3$

$\\ \sf\longmapsto AlCl_3$

$\\ \sf\longmapsto CsBr$

$\\ \sf\longmapsto Ti_2O_4$

$\\ \sf\longmapsto Ag_2S$

4 0

3 years ago

Plz help this is due today

vazorg [7]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers