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2 years ago

How would adding the catalyst nitrogen monoxide (NO) affect this reaction?

Chemistry

1 answer:

RSB [31]2 years ago

8 0

Answer: A) NO increases the rate at which $SO_3$ molecules are formed.

Explanation:

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more reactant molecules can cross the energy barrier by undergoing collisions and convert to products.

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

Thus NO will increase the rate of reaction by lowering the activation energy and thus the colllisions among $SO_2$ and $O_2$ molecules will incraese which in turn will lead to formatioon of more $SO_3$ molecules.

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The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Brrunno [24]

Answer:

The molecular formula of the compound is $C_{8}H_{16}O_{4}$ . The molecular formula is obtained by the following expression shown below

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is $C_{2}H_{4}O$

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = $\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}$

$n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4$

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Molecular formula = 4 $\times$ $C_{2}H_{4}O$

Molecular formula is $C_{8}H_{16}O_{4}$

6 0

2 years ago

SCIENCE - Can someone tell me the ph level of rain water?

amid [387]

Normal rainwater has a pH of 5.6<span> (slightly acidic)</span>

8 0

3 years ago

If 48 g of magnesium reacts with oxygen gas, how many grams of magnesium oxide will be formed according to the following equatio

Aloiza [94]

think it maybe 48 mg0 because there should be 2g magnesium in the word equation

4 0

2 years ago

Read 2 more answers

How many grams of Cl are in a 38.0-g sample of the chlorofluorocarbon CF2Cl2

Katena32 [7]

The molecular formula of chlorofluorocarbon is CF₂Cl₂
Molecular mass of CF₂Cl₂ is 120.9 g/mol
Mass of Cl in 1 mol - 2 x 35.5 = 71 g/mol
in 120.9 g of compound - 71 g of Cl
Therefore in 38 g of sample - 71/120.9 x 38 g of Cl
Mass of Cl in 38 g - 22.31 g of Cl

4 0

3 years ago

A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

icang [17]

Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

$\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C$ .

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

In other words, it takes approximately $0.551\; \rm J$ to raise $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C$ . Therefore, by definition, the specific heat of this material would be approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

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2 years ago