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andrezito [222]
2 years ago
13

3.2 g of a gas at STP occupies a volume of 2.24 L. The gas is

Chemistry
1 answer:
Ronch [10]2 years ago
4 0

Answer:

Oxygen

Explanation:

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Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl
anzhelika [568]

Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]

\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]

\Delta H=-277.4kJ

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0
2 years ago
A biologist working in a lab adds a compound to a solution that contains an enzyme and substrate. This particular compound binds
ad-work [718]

Answer:

Adding more substrate would overcome  the effect of the compound

Explanation:

  • Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
  • Compounds known as inhibitors slow down the rate of enzyme activity.
  • Inhibitors are classified as competitive and non-competitive inhibitors.
  • Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
  • The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
  • Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.

6 0
3 years ago
The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
3 years ago
What does "hidden phenomena" mean in science?
dusya [7]

Answer:

it's means two plural phenomenon and object or aspect known for through the sentence rather than by dot or intuition a temporal or spout temporal object of sensory experience as discussion from a nominal

6 0
3 years ago
What is 5 divided by 2.7​
Strike441 [17]
5 divided by 2.7 is 0.54
3 0
2 years ago
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