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TEA [102]
3 years ago
12

Which process is responsible for changing the composition of rock?

Chemistry
2 answers:
andrezito [222]3 years ago
3 0

Answer:

Chemical Weathering

Explanation:

Took test

qaws [65]3 years ago
3 0

Answer:

It's C on Edge

Explanation:

I just took and passed the test with 100%

You might be interested in
2NaOH + H2So4=______+2H2O<br>​
Dmitry [639]

Answer:

2NaOH + H _{2} SO _{4} = >  \:Na _{2}SO _{4}  +2H _{2} O

7 0
3 years ago
HELP WITH THIS ASAP PLEASE!!! What is the mass in grams of 8.42 x 10^22 atoms of sulfur?
V125BC [204]

Answer:

Answer:

see explanation and punch in the numbers yourself ( will be better for your test)

Explanation:

If you are given atoms you need to divide by Avogadro's number 6.022x10^23

then you will have moles of sulfur-- once you have moles multiply by the molar mass of sulfur to go from moles to grams

mm of sulfur is 32 g/mol

5 0
3 years ago
A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and
kvasek [131]

<u>Given:</u>

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

<u>To determine:</u>

The volume (V) of the balloon

<u>Explanation:</u>

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

Ans: Volume of the balloon is around 488 L


3 0
3 years ago
How many molecules are present in the following quantity<br> 0.250 mole of H2O
klio [65]
If  1mole ------------- is ---------------- 6.02*10²³
than 0.25mole ----- is ----------------  x

x = [0.25mole*6.02*10²³]/1mole = <u>1,505*10²³</u>
4 0
3 years ago
Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Learn more about freezing point here;

brainly.com/question/3121416

#SPJ4

7 0
2 years ago
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