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3 years ago

During a new moon, the moon is

Chemistry

1 answer:

IgorLugansk [536]3 years ago

4 0

It is A. This is because, according to your diagram, the sunlight hits the moon, and it is blocking some of the sunlight to reach the Earth. This means that it is in between both of them.

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A scientist evaluates its livestock and determines which livestock it is

Rina8888 [55]

Any animals look for steangth mostly because the offspring can probobly onther trait is looks. The male will try to impess the female.

4 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

What is the h+ concentration for an aqueous solution with poh = 3.35 at 25 ∘c? express your answer to two significant figures an

RSB [31]

POH value was calculated by the negative logarithm of hydroxide ion concentration.
To know the hydrogen ion concentration, we need to know the pH value, that can be found out if pOH is known
pH + pOH = 14
pH = 14 - pOH
pH = 10.65
once the pH is known we have to find the antilog.
[H⁺] = antilog (-pH)
antilog can be found by
[H⁺] = 10^(-10.65)
[H⁺] = 2.2 x 10⁻¹¹ M

3 0

3 years ago

After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.203 M. You perform

jek_recluse [69]

Answer:

0.203 is the mean of the concentration of the HCl solution

Explanation:

You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:

0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =

<h3>0.203 is the mean of the concentration of the HCl solution</h3>

8 0

3 years ago

Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

$Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}$

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

6 0

2 years ago