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3 years ago

Each molecule of table sugar, C12H22O11, contains

Chemistry

2 answers:

algol [13]3 years ago

7 0

Answer : The correct option is, (d) 12 atoms of carbon.

Explanation:

The formula of given molecule of table sugar is, $C_{12}H_{22}O_{11}$

The given molecule is made up of three elements which are, carbon, hydrogen and oxygen.

In the given molecule of table sugar $C_{12}H_{22}O_{11}$ , there are 12 number of carbon atoms, 22 number of hydrogen atoms and 11 number of oxygen atoms.

From the given options we conclude that, the option (d) is the correct option. While the other options are incorrect.

Hence, the correct option is, (d) 12 atoms of carbon.

Delicious77 [7]3 years ago

3 0

D, because C12 means there's 12 atoms of carbon.

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The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

5. How many moles are in 10g of sodium?

vagabundo [1.1K]

Answer:

0.43

Explanation:

divide the given mass by molar mass from the periodic table

3 0

3 years ago

You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th

ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially, only 4 moles of oxygen gas were present in the flask.

$p_{O_2}=Tp_1\times X_{O_2}$ ( $X_{O_2}=\frac{4}{4}$ ) ( according to Dalton's law of partial pressure)

$p_{O_2}=Tp_1\times 1=Tp_1$ ....(1)

$Tp_1$ = Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

$X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}$

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

$p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

$Tp_2$ = Total pressure of the mixture.

from (1)

$Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

On rearranging, we get:

$Tp_2=2\times Tp_1$

The new total pressure will be twice of initial total pressure.

7 0

3 years ago

What is the mass percent of oxygen (0) in SO2?

vazorg [7]

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

5 0

2 years ago

Compared to oil and gas, solar energy:

prohojiy [21]

Cost more. Gas and oil have a very steady low price, and also the sun is not always out.
Does that help?

6 0

3 years ago

Read 2 more answers