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2 years ago

Match the followings.

Chemistry

1 answer:

insens350 [35]2 years ago

5 0

Answer:

Explanation:

Here, we are to copy and complete the paragraph with the correct options as given below in the question.

Before that, let us have an understanding of what a conversion factor is all about.

A conversion factor refers to a formula that is being used to change over an estimation in a given arrangement of units to its comparable estimation in another arrangement of units. Subsequently, the formula comprises both a number and its unit.

So;

$\text{You can check that you have written the correct conversion factors for an }$

$\text{equality by looking at the numbers and units in each; the information on }$

$\text{one side of the equity appears in the of a conversion factor while the }$

$\text{information on the other side of the equity appears in the denominator.}$

$\text{To generate a second conversion factor, we "turn over" the first }$

$\text{conversion factor so the numbers and units in the numerator and }$

$\text{denominator switch places.}$

You might be interested in

Write equations to show whether the solubility of either of the following is affected by pH: (b) Hg₂(CN)₂.

nata0808 [166]

Increasing pH will increase the solubility of the Hg2(CN)2 by shifting equilibrium to right side.

What is the meaning of OH in chemistry?

The chemical group, ion, or radical OH that consists of one atom of hydrogen and one of oxygen and is neutral or negatively charged.

Hg2(CN)2 + 2OH- ----> 2HgO(s) + 2HCN

adding OH- to the mercury(l) cyanide will cause the formation of the solid HgO.

therefore increasing pH will increase the solubility of the Hg2(CN)2 by shifting equilibrium to right side.

Learn more about OH

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3 0

1 year ago

If these teams are pulling with the same amount of force what will happen?

Usimov [2.4K]

They will not move...

3 0

2 years ago

How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?

Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

S1=50%

V2=250ml

S2= 20%

We all know that

V1S1=V2S2

∴V1= V2×S2÷S1

∴V1= V2S2×1/S1

∴V1= 250×20÷50

∴V1= 100ml

6 0

3 years ago

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 2 atm
T1= 50 C= 323 K (being 0 C= 273 K)
P2= 3.2 atm
T2= ?

Replacing:

$\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}$

Solving:

$T2*\frac{2 atm}{323 K} =3.2 atm$

$T2=3.2 atm*\frac{323 K}{2 atm}$

T2= 516.8 K= 243.8 C

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

5 0

3 years ago

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

The system is not in equilibrium and will evolve left to right to reach equilibrium.

3 0

3 years ago