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3 years ago

X + Y XY + heat What happens as the temperature is increased?

1 answer:

6 0

What happen as temperature is increased is that the equilibrium will shift to the left that is the side of the reactant.

The increase of temperature in is X +Y⇆XY reaction led to the shift of equilibrium to the left (reactants side). This is because the reaction between X and Y is exothermic(release heat) and thus increase in temperature will led to increase in amount of the product. This led to the shift of the balance at equilibrium toward the reactants side.

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The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

Semmy [17]

Answer:

The molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

, $=\frac{0.07}{100}\times 21.5}\\=0.0151g$ of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

$\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M$

The chemical equation for the ionization of copper (II) sulfate follows:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = $4.73\times 10^{-5}M$

Hence, the molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

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

2 years ago

PLEASE PLEASE PLEASE ANSWER HAVE MERCY 35 POINTS

Arada [10]

You should give answers since some people will just try and find any answer, which happened to me. So like if you have answers for the question just put them like A B C D kinda like true or false for others to know what they got instead of giving them more work, and then they just give you random answers.

6 0

2 years ago

Which determines how atoms combine to form molecules?

arlik [135]

Will be letter E cause all of them can combine molecule

7 0

3 years ago

CO2 + H20. What is the total number of moles of Oz required to completely

maria [59]

I think the right answer is 2.5

7 0

3 years ago