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Lunna [17]
3 years ago
12

Chem question!! please help!!!

Chemistry
1 answer:
podryga [215]3 years ago
6 0
Cu is the answer... :)
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What happens to the reaction rate as the reactants get used up in a reaction?
Alex17521 [72]

Answer:

C.  The rate of the reaction begins to decrease as the reactants are used.

Explanation:

a p 3 x

8 0
3 years ago
How does the law of conservation of mass apply to this reaction: C2H4+O2=H2O+Co2?
pav-90 [236]
Consider the formation of water molecule. Hydrogen combines with oxygen to form  a water molecule.

 

In this case, the total mass of the reactants = total mass of the products. Also, the number of atoms of hydrogen and oxygen in the reactants side and the products side are equal.

that's all I know

3 0
3 years ago
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What is the electron configuration for 1s22s22p63s2
patriot [66]

Answer:

its atomic no is 12

Explanation:

name = MAGNESIUM

3 0
1 year ago
The vapor pressure of CCl3F at 300 K is 856 torr. If 11.5 g of CCl F is enclosed in a 1.0-L container, will any liquid be pres-
Lemur [1.5K]

Answer:

Yes, there will be liquid present and the mass is 5.19 g

Explanation:

In order to do this, we need to use the equation of an ideal gas which is:

<em>PV = nRT (1)</em>

<em>Where:</em>

<em>P: Pressure</em>

<em>V: Volume</em>

<em>n: number of moles</em>

<em>R: gas constant</em>

<em>T: Temperature</em>

we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.

From (1), solving for moles we have:

<em>n = PV/RT (2)</em>

Solving for n:

P = 856/760 = 1.13 atm

R = 0.082 L atm / mol K

n = 1.13 * 1 / 0.082 * 300

n = 0.0459 moles

Now, the mass is:

m = n * MM (3)

The molar mass of CCl3F reported is 137.37 g/mol so:

m = 0.0459 * 137.37

m = 6.31 g

Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:

m = 11.5 - 6.31

m = 5.19 g

7 0
4 years ago
Calculate the volume of concentrated 14.8 m ammonia required for the preparation of 3.1 g
Contact [7]
I think it is <span>0.823 g H2O+NH3/1 ml H2O+NH3 17.034 g NH3/1 mol 18.016 g H2O/1 mol 0.87 mol NH3/1 kg H2O</span>
8 0
3 years ago
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