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3 years ago

Which rigid cylinder contains the same number of gas molecules at STP as a 2.0-liter rigid cylinder containing H2(g) at STP?

1 answer:

4 0

When we say STP that means the system is at its standard temperature and pressure. Both systems is having the same condition hence temperature and pressure is constant. We then set the amount of gas in both cases to be equal. Hence from the ideal gas law,
PV=nRT we can say that with everything constant, volume must also be constant.

The answer is (2) 2 L of methane gas.

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Mendeleev created this table as he noticed that a

Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements whose properties were similar to the properties of the noble gases. This would have lead him to suspect their existence.

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3 years ago

A story on chemical reaction or electric current

Anuta_ua [19.1K]

Answer:

a process that involves rearrangement of the molecular or ionic structure of a substance, as opposed to a change in physical form or a nuclear reaction

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3 years ago

Read 2 more answers

How many grams of oxygen gas occupy 12.3 L of space at 109.4 kPa and 15.4oC?

I am Lyosha [343]

17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of the gas must be calculated first as follows:

PV = nRT

Where;

P = pressure = 1.0796941atm
V = volume = 12.3L
n = number of moles
T = temperature = 288.4K
R = gas law constant = 0.0821 Latm/molK

1.079 × 12.3 = n × 0.0821 × 288.4

13.27 = 23.68n

n = 13.27/23.68

n = 0.56mol

Mass = 0.56 × 32

mass of oxygen gas = 17.93g

Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.

Learn more about mass at: brainly.com/question/19694949

3 0

2 years ago

How much energy does it take to boil 100 mL of water?

yawa3891 [41]

Answer:

the answer is D. Because the 1 atm pressure of water is 40.65 or 40.7.

8 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago