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Irina-Kira [14]
3 years ago
7

An electron has a charge identical to that of

Chemistry
1 answer:
kenny6666 [7]3 years ago
3 0
Protons! I hope that helped out!
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Which variables does equilibrium depend on?
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I believe tempurature, though i'm not sure. Give it a shot and tell me what you got!

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I'll be giving out brainliest to whoever chooses more than one!!!
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I think it might be B, and A

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What volume of 10M sulfuric acid is needed to create 2L solution of<br> 7.5M?
Anvisha [2.4K]

Explanation: the answer to life the universe and everything is 42.

School doesnt matter

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8 0
3 years ago
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha
lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

\frac{60}{100}\times 3.6 = 2.16 L

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = \frac{50}{100}\times 3.6=1.8 L

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

2.16 L-0.6x=1.8 L

x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0
4 years ago
A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori
Yuliya22 [10]

Answer:

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

5 0
3 years ago
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