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Aleks [24]
4 years ago
5

A 5.0 l sample of gas is cooled from 298 k to 278 k at constant pressure. what is the new volume of the gas?

Chemistry
1 answer:
olga nikolaevna [1]4 years ago
5 0
In order to calculate the final volume of the gas, we may apply Charles Law, which states that for a fixed amount of gas subject to a fixed amount of pressure, the volume and temperature of a gas are directly proportional. This is:
V/T = constant
Using this equation,
5/298 = V/278
V = 4.66 L

The final volume of the gas is 4.66 liters
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8 0
3 years ago
Round the following numbers to the number of significant figures indicated:
Kruka [31]
  1. 3.5
  2. 9
  3. 24.7
  4. 8.01
  5. 0.006
  6. 13(1.25×1012=12.65)

Explanation:

Above we have simply looked the condition that where we put decimal point that gives your needed answer after round off.

For example:

we have to make 2 s.f from 345 so we kept decimal after 3 and we rounded off to make 3.5 from 3.45

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7 0
3 years ago
What is the result of fusion reactions?
Shtirlitz [24]
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3 0
3 years ago
Read 2 more answers
(f) Which of the following alkenes is the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol?
belka [17]

Answer:

D) 2-methylpent-2-ene

Explanation:

This is an elimination reaction of Halogenoalkane. 2-bromo-2-methylpentane when is heated with NaOH or NaOC2O5( sodium ethoxide) in ethanol will form alkene rather than alcohol.

2-methylpent-1-ene is minor product since double bond form with secondary Carbon rather than primary Carbon.

8 0
3 years ago
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An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was
Yuri [45]

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

<em>Moles HCl added:</em>

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

<em>Moles NaOH to titrate the excess:</em>

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

<em>Moles of acid that were neutralized:</em>

0.0105 moles - 0.0010 moles =

<h3>0.0095 moles of acid were neutralized by the antiacid</h3>
7 0
3 years ago
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