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Murrr4er [49]
2 years ago
7

When the reaction begins, the researcher records that the rate of reaction is such that 1 mole of A is consumed per minute. Afte

r making changes to the reaction, the researcher notes that 2 moles of A are consumed per minute. What change could the researcher have made to effect this change?
A. Utilize reaction conditions to convert both reactants to solids.
B. Decrease the concentration of reactant B to allow C to be produced at a greater rate.
C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy.
D. Increase the concentration of C to allow for more frequent collisions of A and B of higher energy
E. Introduce a catalyst to decrease the consumption of A and B.
Chemistry
1 answer:
yanalaym [24]2 years ago
8 0

Answer:

C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy.

Explanation:

For the reaction:

A + B → 2C

You can increase the consume of A:

A. Utilize reaction conditions to convert both reactants to solids.  <em>FALSE. </em>Convert both reactants in solids makes less frequent collisions doing difficult the reaction.

B. Decrease the concentration of reactant B to allow C to be produced at a greater rate.  <em>FALSE. </em>The decreasing of B will make the reaction slowly.

C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy.  <em>TRUE. </em>The increase of temperature will make more frequent collisions of A and B doing the reaction faster.

D. Increase the concentration of C to allow for more frequent collisions of A and B of higher energy . <em>FALSE. </em>The increase of C makes less frequent collisions doing difficult the reaction.

E. Introduce a catalyst to decrease the consumption of A and B. <em>FALSE. </em>Introducing a catalyst increase the consumption of A and B.

I hope it helps!

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Which of the following statements about monosaccharide structure is true?
Nana76 [90]

Answer:

The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)

Explanation:

Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5

8 0
2 years ago
The Rydberg constant is 3.3 x 10 15 Hz. this value is also
mina [271]

Answer:

Rydberg constant 3.3 x 10¹⁵ Hertz is equal to 1.090 x 10⁷ m⁻¹

Explanation:

Given;

Rydberg constant as  3.3 x 10¹⁵ Hz

1 Rydberg constant = 3.3 x 10¹⁵ Hz

1 Rydberg constant  = 1.090 x 10⁷ m⁻¹

Therefore, Rydberg constant 3.3 x 10¹⁵ Hertz is equal to 1.090 x 10⁷ m⁻¹

7 0
2 years ago
Read 2 more answers
Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion.
raketka [301]

Explanation:

The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.

This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose H^{+} ion.

Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.

Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.

       HOI < HOBr_{2} < HOCl_{3} < HOF

8 0
2 years ago
2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b
Virty [35]

Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: 2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl

     Compound                                 Molar mass (g/mol)

         NaOH                                           39.997

         MgCl_{2}                                           95.211

        Mg(OH)_{2}                                        58.3197

So, 2.50 g of NaOH = \frac{2.50}{39.997} mol of NaOH = 0.0625 mol of NaOH

      4.30 g of MgCl_{2}  = \frac{4.30}{95.211} mol of MgCl_{2} = 0.0452 mol of MgCl_{2}

According to balanced equation-

2 mol of NaOH produce 1 mol of Mg(OH)_{2}    

So, 0.0625 mol of NaOH produce (\frac{0.0625}{2}) mol of NaOH or 0.03125 mol of NaOH

1 mol of MgCl_{2} produces 1 mol of Mg(OH)_{2}

So, 0.0452 mol of MgCl_{2} produce 0.0452 mol of Mg(OH)_{2}

As least number of moles of Mg(OH)_{2} are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of Mg(OH)_{2} would be produced = 0.03125 mol

                                                                           = (0.03125\times 58.3197) g

                                                                           = 1.822 g

6 0
2 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
2 years ago
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