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3 years ago

How are igneous rocks classified?

Chemistry

2 answers:

ivolga24 [154]3 years ago

5 0

I think it would be answer 1.

ziro4ka [17]3 years ago

3 0

Pretty sure the answer is 1

You might be interested in

If I have 22,9 moles of a gas at a pressure of 12.4 atm and a volume of 24 liters, what is the temperature?

Grace [21]

Answer:

160 K, using 2 SF,

Explanation:

Ideal Gas law, rearrange to solve for T (K), answer about 158.3 round to 2SF

PV = nRT

8 0

2 years ago

Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.

Naya [18.7K]

Answer:

The answer to your question is: 8.82 x 10 ²⁴ atoms of oxygen

Explanation:

Data

Cube measuring : 2.805in on each edge

density = 2.7 g/cm3

# of oxygen atoms in the cube = ?

Process

Volume of the cube = (2.805)³ = 22.07 in³

Convert in³ to cm³ 1 in³ ------------------- 16.39 cm³

22.07 in³ -------------------- x

x = (22.07 x 16.39) / 1 = 361.72 cm³

Density = mass / volume

Mass = density x volume

Mass = (2.7)(361.72) = 976.66 g

Atomic mass CaCO3 = 40 + 12 + (16 x 3) = 100 g

100 g of CaCO3 -------------------- 48 g of O2

976.66 g of CaCO3 --------------- x

x = (976.66 x 48) / 100 = 468.8 g of O2

32 g of O2 ---------------------- 1 mol

468.8 g of O2 -------------------- x

x = (468.8 x 1) / 32 = 14.65 mol

1 mol ----------------------- 6.023 x 10 ²³ atoms

14.65 mol ------------------- x

x = (14.65 x 6.023 x 10 ²³) / 1 = 8.82 x 10 ²⁴ atoms of oxygen

3 0

3 years ago

SICL4 + H2O sio2 hcl

atroni [7]

Do you want this equation balanced?

If so, it would be

SiCl4 + 2H2O ---> SiO2 + 4HCl

5 0

3 years ago

A gas is contained in a thick-walled balloon. When the pressure changes from 39.1 bar to 87.0 bar the volume changes from blank

lidiya [134]

Answer: -

1.34 L

Explanation: -

Initial Pressure P 1 = 39.1 bar

Initial Temperature T 1 = 643 K

Let the initial volume be V 1.

Final pressure P 2 = 87.0 bar

Final temperature T 2 = 525 K.

Final volume V 2 = 0.492 L

Using the equation

$\frac{P1 V 1}{T1} = \frac{P 2 V 2}{T2}$

$V 1 = \frac{P2 V2 T1}{P 1 T2}$

Plugging in the values

We have

V 1 = 87 bar x 0.492 L x 643 K / (39.1 bar x 525 K)

= 1.34 L

Thus, a gas is contained in a thick-walled balloon. When the pressure changes from 39.1 bar to 87.0 bar the volume changes from 1.34 L to 0.492L and the temperature changes from 643K to 525K

4 0

4 years ago

Read 2 more answers

Group 18 noble gases are relatively inert because *

choli [55]

Answer:I believe its A. There noble gasses so there happy with there electron count!

Explanation:

3 0

3 years ago