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2 years ago

In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Chemistry

1 answer:

Oksanka [162]2 years ago

8 0

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In $H_2O$ , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in $Cl^-$ is -1. On product side, the oxidation number of hydrogen in $H_2$ is zero and in $OH^-$ the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in $Cl_2$ is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

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Todd drives west to work;54km on way in 1.5hours and then drives home in 2hrs. What is his velocity on the way to work? What is

Nataly [62]

Answer:

Explanation:

Velocity is a physical quantity that expresses the relationship between the space traveled by an object, the time taken for it and its direction. In other words, velocity is associated with the change of position of a body over time.

Its unit in the International System of Units is the meter per second ( $\frac{m}{s}$ ), but it can also be expressed in $\frac{km}{hr}$

So, in this case, the speed on the way to work will be:

$velocity=\frac{54 km}{1.5 hours}$

velocity= 36 $\frac{km}{hr}$

The velocity on the way back home will be:

$velocity=\frac{54 km}{2 hours}$

velocity= 27 $\frac{km}{hr}$

7 0

3 years ago

Read 2 more answers

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc

max2010maxim [7]

Answer: The percent yield of the water is 31.98 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

For oxygen gas:

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

4 0

3 years ago

Standard notation for 31,254,000

bearhunter [10]

That seems to be standard notation

4 0

3 years ago

What is the boiling point of methanol, if ∆H°for vaporization is 35.3 kJ/mol, and ∆S° for vaporization is 105 J/mol K?

QveST [7]

Answer: 336.2K

Explanation:

The boiling point is obtained from the standard enthalpy and entropy of vaporization as shown in the image attached.

Standard enthalpy of vaporization=35.3KJmol-1=35.3×10^3 Jmol-1

Standard entropy of vaporization= 105 JMol-1K-1

The formula and details of the solution are shown in the image attached.

6 0

3 years ago

A quantity of 7.480 g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pre

Serga [27]

Answer:

C5H6N2O3

Explanation:

First the empirical formulas

C= 41.8÷12= 2.48

H= 4.7÷1= 4.7

O= 37.3÷ 16= 2.33

N= 16.3÷14 = 1.16

Divide by the smallest

C= 3.48/1.16=3

H= 4.7/1.16=4.1

O= 2.33/1.16=2

N= 1.16/1.16=1

Therefore empirical formula = C3H4NO2

To calculate molecular formula for osmotic pressure,

π= cRT

π=cgRT/M where cg is in g/liter & T is temperature in Kelvin. Thus

π= (7.480*0.0821*300)/ M

M= 184.23/1.43

M= 128.83

To find molecular Formula

Molecular Mass= (empirical mass)n

128.83= (C3H4NO2)n

128.83= 86n

n= 1.5

Therefore the molecular formula

(C3H4NO2)1.5

= C4.5H6N1.5O3

Approximately

C5H6N2O3

7 0

3 years ago