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1 year ago

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The equilibrium reaction below has the Kc = 4.00 at 25°C. If the temperature of the system at equilibrium is increased to 100°C,

how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change.

1 answer:

professor190 [17]1 year ago

6 0

Answer:

Following are the answer to this question:

Explanation:

In the given question information is missing, that is equation which can be defined as follows:

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g) \ \ \bigtraingleup H^0=+206.2KJ$

Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right

Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient Kc = \frac{product}{reactant} It increases [product] but reduces [reactant] Therefore, Kc increases

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Liquid Q is a polar solvent and liquid R is a nonpolar solvent. On the basis of this information, you would expect:

bearhunter [10]

4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."

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2 years ago

In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made

LenKa [72]

The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>

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1 year ago

A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?

rodikova [14]

Answer:

$\boxed {\boxed {\sf D. \ 64 \ grams }}$

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

$\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

Multiply by the given number of moles, which is 2.0

$2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

The moles of methanol will cancel each other out.

$2.0 \ *\frac{32 \ g \ CH_3OH}{1 }$

The denominator of 1 can be ignored.

$2.0 * 32 \ g\ CH_3OH$

Multiply.

$64 \ g \ CH_3OH$

There are 64 grams of methanol in the sample.

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1 year ago

A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

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1 year ago

How many hydrogen bonds can CH2O make to water

VladimirAG [237]

Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:

<span>H2C=O---------H-OH </span>

<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>

<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>

<span>H2C=O + H2O <-----------------> H2C(OH)2</span>

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1 year ago