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enyata [817]
3 years ago
15

Solving fractions of

Chemistry
2 answers:
Ilya [14]3 years ago
7 0

1) 1/2 of 24= 12

2) 1/3 of 24= 8

3) 1/4 of 24= 6

hope this helped

LuckyWell [14K]3 years ago
6 0
1.is 9
2.is 4oz.
There you go
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Is it possible for a reaction to have a large equilibrium constant but small forward and reverse rate constants?
emmainna [20.7K]

The equilibrium constant K and the forward rate constant k1 and backward rate constant k2 has the following relation:

K = k1 / k2

 

So from the equation, we can say that yes it is possible to have large K even if k1 is small given that k2 is very small compared to k1: (k2 very less than 1)

<span>k2 << k1</span>

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4 years ago
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Given the following proposed mechanism, predict the rate law for the overall reaction. 2 NO2 + Cl2 → 2 NO2Cl (overall reaction)
alex41 [277]

Answer:

Explanation: the module will end up working out when you mix NO2CI and CI

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3 years ago
Which element at stp is a liquid that conducts electricity well?
Slav-nsk [51]
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If the density of water is 1.00 g/mL, will an object with a density of 3.4 g/mL sink or float?
miss Akunina [59]

Explanation:

Objects or substances with their density greater than that of water will sink in it whiles those less than water will float on it.

From the question the object has a density of 3.4 g/mL.

Since it's density is greater than that of water the object will sink.

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8 0
4 years ago
20 L of nitrogen gas are collected at a temperature of 50°C and 2 atm. How many grams of nitrogen gas were collected?
Andreas93 [3]

Answer:

0.1077 grams

Explanation:

First we will employ the ideal gas law to determine the number of moles of nitrogen gas.

PV=nRT

P=2 atm

V=20L

R=0.08206*L*atm*mol^-1*K^-1

T=323.15 K

Thus, 2atm*20L=n*0.08206*L*atm*mol^-1*K^-1*323.15K

K, atm, and L cancels out. Thus n=2*20mol/0.08206*323.15=1.5 moles

Lastly, we must convert the number of moles to grams. This can be done by dividing the number of moles by the molar mass of nitrogen gas, which is 14 grams.

1.5/14=0.1077 grams

8 0
3 years ago
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