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3 years ago

Calculate number of CO2 molecules in 0.250 mol of co2

1 answer:

8 0

Carbon's molar mass is 12gm

Oxygen's molar mass is 16gm

So CO2 molar mass is 44gm

Hence 1mol of CO2 contains 44gm
There 1/4mol of CO2 contains 44/4=11gm.

Hope it helps

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A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

sleet_krkn [62]

Answer:

This is under gas laws. check it out

3 0

2 years ago

Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.7kJ$ ( × 2)

(2) $H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_2=-285.9kJ$ ( × 2)

(3) $2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_3=-1411kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

6 0

3 years ago

HURRY PLEAS!!!!!!!!!!!!!!!!!!!!

mixas84 [53]

D. action force and reaction forces are stated in his law

8 0

2 years ago

Read 2 more answers

Imagine that you are observing an enzyme-catalyzed reaction in lab. every time you add more enzyme, the reaction rate increases

tester [92]

Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.

6 0

2 years ago

What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O

nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

Concentration of the base (Cb): 0.300 M

Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

$[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M$

Step 4: Calculate the pOH

We will use the following expression.

$pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6$

Step 5: Calculate the pH

We will use the following expression.

$pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4$

8 0

2 years ago