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vaieri [72.5K]
3 years ago
14

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00- mL sam

ple of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium.
Chemistry
1 answer:
noname [10]3 years ago
8 0

Answer:

Concentration ammonium is 0.036 g/mL

Explanation:

Stock solution has concentration of 0.108g/mL. Take 10mL stock solution and add 50 mL of water meaning it has 1,08g/60mL. Ammonium sulfate has 2 molecules of cation ammonium cation. So, (1,08g/60mL.)*2=0.036 g/mL

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An object that is 0.5 m above the ground has the same amount of potential energy as a spring that is stretched 0.5 m. Each dista
Snowcat [4.5K]

Answer: The correct answer is The aplastic potential energy of the spring will be two times greater than the gravitational potential energy of the object.

Explanation: The formula for Gravitational potential energy is= mgh where

m= mass

g= 9.8

h= height

On the other hand the formula for elastic potential energy is (1/2)KX^2

Where K is the spring. By changing the values of H and X, we will see elastic potential energy will remain more.

7 0
3 years ago
Read 2 more answers
a 25 ml sample of gas is in a syringe at 22C if it was cool down to 0 C what will the volume of the gas become
Tpy6a [65]

Answer: 0.023 liters

Explanation:

Given that,

Original volume of gas (V1) = 25mL

[convert 25mL to liters

If 1000ml = 1L

25ml = 25/1000 = 0.025L]

Original temperature of gas (T1) = 22°C

[Convert 22°C to Kelvin by adding 273

22°C + 273 = 295K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 0°C

[Convert 0°C to Kelvin by adding 273

0°C + 273 = 273K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

0.025L/295K = V2/273K

To get the value of V2, cross multiply

0.025L x 273K = 295K x V2

6.825L•K = 295K•V2

Divide both sides by 295K

6.825L•K/295K = 295K•V2/295K

0.023 L = V2

Thus, the new volume of the gas will be 0.023 litres

6 0
4 years ago
When a solution of NaI reacts with a solution of AgNO3, what is the net ionic equation?
Leni [432]

Answer:

Ag⁺(aq) + I⁻(aq) → AgI(s)

Explanation:

Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.

In the reaction:

AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).

The ionic equation is:

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).

Now, listing only the species that are participating in the reaction:

<h3>Ag⁺(aq) + I⁻(aq) → AgI(s)</h3>

3 0
4 years ago
Fill in the blanks.
Maru [420]

Answer:

Rhombic sulphur and monoclinic sulphur.

5 0
3 years ago
Given: N2 + 3Cl2 → 2NCl3, ΔH = 464 kJ/mol Use the given bond energies and the periodic table to calculate the energy change in t
guapka [62]

The given chemical reaction is:

N_{2}+3Cl_{2}-->2NCl_{3}

ΔH^{0}_{reaction}=∑BE(reactants)-∑BE(products)

                 = {(941 kJ/mol) + (3 * 242 kJ/mol)} -[{2*(3*200 kJ/mol)}]

                     = 467 kJ/mol

Calculating the change in heat when 85.3 g chlorine reacts in the above reaction:

Moles of chlorine = 85.3 g Cl_{2}* \frac{1 mol Cl_{2}} {70.91 g Cl_{2}  }

                             = 1.20 mol Cl_{2}

Heat change when 1.20 mol chlorine reacts

                             = 1.20 mol * \frac{467kJ}{mol} =560.4 kJ

6 0
3 years ago
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