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vladimir1956 [14]
3 years ago
11

In a rigid container, when the speed of the gas molecules decreases, the pressure of the gas also decreases. true or false

Chemistry
1 answer:
KIM [24]3 years ago
8 0
In order to understand this question and to answer it correctly, there are some definitions/terms that need to be acknowledged.

 Firstly, TEMPERATURE is defined as the average KINETIC ENERGY of a sample. So, in the question when the reference is made to the speed of the molecules, it can be translated to temperature. Thus, since the speed of the molecules decreases, then it can be said that the Temperature of the system decreases.

Secondly, according to Gay-Lussac Law, Pressure is directly proportional to Temperature when the volume is constant. As such, an increase in Pressure sees an increase in Temperature.

Based on what is known above, the statement "<span>In a rigid container, when the speed of the gas molecules decreases, the pressure of the gas also decreases" is a true one.

Thus the answer is TRUE.</span>
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I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
4 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
3 years ago
Can you guys give some science fair project ideas(8th grade)
Anton [14]
Answer
i’m in 7th grade but for this years science fair i did, “ how does temperature affect the elasticity of rubber bands”

Guide a growing plant through a maze.
8th Grade Science Plant Maze KiwiCo

Prove that plants really do seek out the light by setting up a simple or complex maze. This is a simple 8th grade science project with really cool results.

Blow out a candle with a balloon.


Blowing up a balloon with baking soda and vinegar is the classic acids and bases experiment. Take it a step further by experimenting with the carbon dioxide it produces. (Don’t be afraid of fire in the science

Stand on a pile of paper cups.


Combine physics and engineering and challenge 8th grade science students to create a paper cup structure that can support their weight. This is a cool project for aspiring architects.

hope this helps and have a wonderful day :)
5 0
3 years ago
Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid
jasenka [17]

Answer: a) The K_a of acetic acid at 25^0C is 1.82\times 10^{-5}

b) The percent dissociation for the solution is 4.27\times 10^{-3}

Explanation:

CH_3COOH\rightarrow CH_3COO^-H^+

 cM              0             0

c-c\alpha        c\alpha          c\alpha

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.10 M and \alpha = ?

Also pH=-log[H^+]

2.87=-log[H^+]  

[H^+]=1.35\times 10^{-3}M

[CH_3COO^-]=1.35\times 10^{-3}M

[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M

Putting in the values we get:

K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}

K_a=1.82\times 10^{-5}

b)  \alpha=\sqrt\frac{K_a}{c}

\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}

\alpha=4.27\times 10^{-5}

\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}

5 0
3 years ago
Look at the image shown. What does this image represent?
JulijaS [17]

Answer:

Linear molecule with two domains

Explanation:

8 0
3 years ago
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