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3 years ago

What happens to pressure as temperature increases and temperature decreases

Chemistry

2 answers:

likoan [24]3 years ago

8 0

When temperature increases pressure also increases.

When temperature decreases pressure decreases.

That is why you have low tire pressure when it’s cold out

Tasya [4]3 years ago

3 0

Temperature and Pressure have a <u>direct relationship</u>!

When Temperature increases ↑

Pressure increases ↑

When Temperature decreases ↓

Pressure decreases ↓

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EXPERIMENT 4 : TYPES OF PLATE BOUNDARY

zalisa [80]

Answer:

Movement in narrow zones along plate boundaries causes most earthquakes. Most seismic activity occurs at three types of plate boundaries—divergent, convergent, and transform. As the plates move past each other, they sometimes get caught and pressure builds up.

Explanation:

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2 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

Hello guys can you help me on this.

RideAnS [48]

Answer:

1.Sulfur dioxide and nitrogen oxide

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3 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

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2 years ago

An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well Exponents in rate laws are

Karo-lina-s [1.5K]

Answer:

- False.

- True.

Explanation:

Hello, for each statement we state:

- An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well.

FALSE because considering a rate law like:

$-r=kC^2$

The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

$-r=kC$

The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

- Exponents in rate laws are based on the coefficients from the balanced equation.

FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

- The rate constant, k, takes into account the effect of activation energy and temperature on the reaction.

TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

$k=Aexp(-\frac{Ea}{RT})$

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

$\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...$

Best regards.

5 0

2 years ago