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3 years ago

ii. Use the solubility table to label the state symbols of each compound in the reaction. (1 point) AgNO3 + KCL = AgCl + KNO3

Chemistry

1 answer:

rjkz [21]3 years ago

4 0

Answer:

AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)

Explanation:

In a solubility table you find:

AgNO₃ (silver nitrate) is highly soluble

KCl (potassium chloride) is soluble

AgCl (silver chloride) is insoluble

KNO₃ (potassium nitrate) is soluble

In a chemical equation the states of soluble compounds is identified as aqeous, using the letter "aq" in parenthesis, and the state of insoluble compounds is identified as solid, using "s" in parenthesis.

Then, the reaction showing the states of the reactants and products is:

AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)

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In Niels Bhor's model of the atom, electrons move

exis [7]

Bohrs model says that electrons move in fixed shells (which have fixed distances) around the nucleus of an atom.

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3 years ago

What is the concentration of OH − and pOH in a 0.00066 M solution of Ba ( OH ) 2 at 25 ∘ C? Assume complete dissociation.

Allushta [10]

Answer: The hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

Explanation:

We are given:

Concentration of barium hydroxide = 0.00066 M

The chemical equation for the dissociation of barium hydroxide follows:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution

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

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M$

Putting values in above equation, we get:

$pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88$

Hence, the hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

3 0

3 years ago

In the laboratory you dilute 5.32 mL of a concentrated 6.00 M perchloric acid solution to a total volume of 100 mL. What is the

8_murik_8 [283]

Answer:

0.3192 M

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5.32 mL Molarity of stock solution (M1) = 6 M

Volume of diluted solution (V2) = 100 mL

Molarity of diluted solution (M2) =?

We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:

M1V1 = M2V2

6 × 5.32 = M2 ×100

31.92 = M2 × 100

Divide both side by 100

M2 = 31.92 / 100

M2 = 0.3192 M

Therefore, the molarity of the diluted solution is 0.3192 M.

7 0

3 years ago

A helium-filled weather balloon has a volume of 512 L at 18.9°C and 756 mmHg. It is released and rises to an altitude of 2.14 km

vazorg [7]

633.97 L

Explanation:

Well use the combined gas law;

P₁V₁T₁ = P₂V₂T₂

We need to change the temperatures into Kelvin;

18.9°C= 292.05 K

5.9°C = 279.05 K

756 * 512 * 292.05 = 639 * V₂ * 279.05

113,044,377.6 = 178,312.95 V₂

V₂ = 113,044,377.6 / 178,312.95

V₂ = 633.97 L

3 0

3 years ago

Read 2 more answers

Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3

wariber [46]

Answer:

More H2(g) is added to the container : Towards products.

CO is removed from the container : Towards reactants.

More CH4(g) is added to the container : Towards reactants

H2O(g) is removed from the container : Towards products.

The contents of the container are heated up. : Towards the reactants.

The contents of the container are cooled down : Towards the products.

The pressure inside the container is increased. :Towards the products

The container is stretched to increase the volume: Towards the reactants.

Explanation: :

CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

There is released heat, so this reaction is exothermic

If the H2 concentration is increased, the system will try to change the concentration change by shifting the balance to the right, and thus the concentration of products will increase. Towards products.

If the CO is removed, the system will try to change this situation by shifting the balance to the left, and thus the concentration of reactants will increase, the concentration of products will decrease. Towards reactants.

If the CH4 concentration is increased, the system will try to change the concentration change by shifting the balance to the left, and thus the concentration of reactants will increase. Towards reactants

If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. Towards products.

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. Towards the reactants.

This because the extra heat / energy must be used.

If the temperature is decreased, the system will produce more heat So the balance will shift to the right. Towards the products.

This because more heat /energy needs to be produced to make up for the loss of heat (energy).

If the pressure is increased, the system will shift to the side with fewer moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the right. Towards the products. An increase of pressure has the same effect on the equilibrium as a decrease of the volume.

If the volume is increased, this means the pressure is decreased, the system will shift to the side with most moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the left. Towards the reactants. An increase of volume has the same effect on the equilibrium as a decrease of the pressure.

6 0

3 years ago