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Olegator [25]
2 years ago
14

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

Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
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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In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so
blsea [12.9K]

Answer:

Concentration of H3O⁺  [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO =  0.3397 g

number of  moles of MO = (mass number of MO) / (molar mass of MO)

number of  moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol  /  19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺  [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺  [H3O⁺] = 0.864 M

5 0
2 years ago
Calculate the number of kilojoules of energy required to convert 50.0 grams of solid DMSO initially at a temperature of 19.0°C t
GuDViN [60]

Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

∆T = change in temperature

According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

Q = 90 × 231

Q = 20790 Joules

To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

Hence, 20.79 kilojoules of energy is required to convert 50.0 grams of solid DMSO to gas.

4 0
2 years ago
When 0.40 mol Al is mixed with 0.40 mol Br2, the following reaction occurs: 2Al(s) + 3Br2(l) → 2AlBr3(s) Identify the limiting r
Lera25 [3.4K]

Answer:

d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Explanation:

From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)

2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3

Hence 1 mole of Br combines with 2/3 moles of Al

or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving

0.4 - 0.267 = 0.133 moles of Al remaining unreacted

7 0
2 years ago
7. Two teams are competing in a tug-of-war contest. Team A is pulling at 4000N and Team B is pulling at 4900N is the opposite di
Leto [7]
Basically team B would win since it is exerting a force of 900N unlike team A ( you can tell by doing 4900N minus 4000N ). It is very unbalanced.

5 0
2 years ago
Read 2 more answers
A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.
enot [183]

Answer:

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

Explanation:

<u>Step 1:</u> Data given

A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K

Moles of CO2 = 1.27 moles

Moles of CO = 3.04 moles

Moles of Ar = 1.50 moles

<u>Step 2:</u> Calculate total number of moles

Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles

<u>Step 3:</u> Calculate mol fraction Ar

Mol fraction Ar = 1.50 mol/5.81 mol = 0.258

<u>Step 4</u>: Calculate partial pressure

1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

8 0
2 years ago
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