It is possible to increase a reaction rate by adding very small amounts of a catalyst, and this is because...

Each of the following phrases matches one of the models below. Match the statement with its appropriate model.

muminat

Not sure about any others but I believe 3 goes with B

4 0

1 year ago

A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is sub

stepladder [879]

<u>Answer:</u> The new water level of the cylinder is 24.16 mL

<u>Explanation:</u>

To calculate the volume of water displaced by silver, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

$10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL$

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

5 0

3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago

Can a solution with undissolved solute be supersaturated

Helen [10]

Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.

Explanation:

A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.

According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.

5 0

3 years ago

The reaction takes place in water. What happens to the equilibrium when the pressure is increased? a)It favors formation of prod

OLga [1]

Answer:

I don't really get the options but it favoures the reactant side.

Explanation:

Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g

If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.

With the explanations I have made, I hope the question is now clear to you.

6 0

4 years ago