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3 years ago

The sphere is the layer of the earth that contains a mixture of gases

Chemistry

2 answers:

jasenka [17]3 years ago

6 0

Omoze layer I know this

konstantin123 [22]3 years ago

6 0

No... I believe that the part of Earth that contains mixtures of gases is the atmosphere, along with its layers.

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Plz answer thx<br><br><br> How many moles are in 2.5 x 10^18 grams of water?

alex41 [277]

Answer:

$1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

Explanation:

Mass of water = $m = 2.5\times 10^{18} g$

Molar mass of water = M = 18 g/mol

Moles = n = $\frac{m}{M}$

$n=\frac{2.5\times 10^{18} g}{18 g/mol}=1.39\times 10^{17} moles$

So, there are $1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

3 0

3 years ago

Show work, thanks

GarryVolchara [31]

Answer:

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution

7 0

2 years ago

Does apple juice have acid

stepladder [879]

Yes, it has many natural acids. However, the biggest and most prominent acid is the organic acid, malic acid.

8 0

3 years ago

How many electrons does phosphorous (P) need to gain to have a stable outer electron level?

Brut [27]

The correct answer is 3.

4 0

3 years ago

Read 2 more answers

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago