Someone help asap please i suck at chem

A mixture of NaBrO 3 , NaBrO3, NaHCO 3 , NaHCO3, Na 2 CO 3 , Na2CO3, and NaBr NaBr was heated, producing H 2 O , H2O, CO 2 , CO2

nata0808 [166]

Answer:

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

Explanation:

For the reactions:

2NaBrO₃(s) ⟶ 2NaBr(s) + 3O₂(g)

2NaHCO₃(s) ⟶ Na₂O(s) + H₂O(g) + 2CO₂(g)

Na₂CO₃(s) ⟶ Na₂O(s)+CO₂(g)

All H₂O(g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

1.83g H₂O ₓ (1 mol H₂O / 18.02g) ₓ (2 mol NaHCO₃ / 1 mol H₂O) = 0.203moles NaHCO₃ ₓ (84g / 1mol NaHCO₃) =

17.06g of NaHCO₃

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

15.51g CO₂ ₓ (1mol / 44.01g) = 0.352moles of CO₂

As 2 moles of NaHCO₃ produce 2 moles of CO₂, moles of CO₂ that comes from NaHCO₃ are 0.203moles NaHCO₃. Moles of CO₂ that comes from Na₂CO₃ are:

0.352mol CO₂ - 0.203mol CO₂ = 0.149mol CO₂

These moles of CO₂ are produced from:

0.149mol CO₂ ₓ (1 mol Na₂CO₃ / 1 mol CO₂) ₓ (106g / 1mol Na₂CO₃) =

15.84g of Na₂CO₃

And all O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

2.87g O₂ ₓ (1 mol O₂ / 32g) ₓ (2 mol NaBrO₃ / 3 mol O₂) ₓ (150.9g / 1mol NaBrO₃) =

9.02g of NaBrO₃

If initial mass of the mixture was 124.5g, mass of NaBr was:

124.5g - 9.02g of NaBrO₃ - 15.84g of Na₂CO₃ - 17.06g of NaHCO₃ =

82.58g NaBr

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

5 0

4 years ago

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

Based on the kinetic molecular theory, which of the following statements is correct about the particles in a sample of gas at a

fgiga [73]

Answer:

There is a lot of empty space between them.

Explanation:

The kinetic molecular theory postulates that a substance is made up of tiny particles called molecules. The molecules of a gas are in constant random motion and collide elastically with each other. They also collide with the walls of the container.

The magnitude of intermolecular forces of attraction between gas molecules is very small. Hence gas molecules are largely apart with a lot of empty space between gas molecules.

8 0

3 years ago

The mass of a mole of carbon atoms is 12.011 grams. What is the mass of a single atom of carbon?

mixer [17]

Amount of substance can be measured in the number of moles.
Avagadros constant states that 1 mol of any substance is made of 6.022 x 10^23 units.
These units could be atoms making up elements or molecules making up compounds
1 mol of C is made of 6.022 x 10^23 atoms of C
Since 1 mol of C has a mass of 12.011 g
Therefore 6.022 x 10^23 atoms of C weigh 12.011 g
Then 1 atom of C weighs -12.011 g / 6.022 x 10^23 atoms = 1.99 x 10^(-23) g
Mass of 1 C atom is 1.99 x 10^(-23) g

8 0

3 years ago

What components of blood can be examined

olganol [36]

Plasma or white blood cells
..........

3 0

3 years ago