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Sophie [7]
3 years ago
11

How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)

Chemistry
1 answer:
nikdorinn [45]3 years ago
8 0

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200.  * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

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+2B\rightleftharpoons C

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 trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Write the following number using Scientific notation : 13,215,296.50 *​
ludmilkaskok [199]

Answer:

13.22x10^7

Explanation:

7 0
3 years ago
Measure out 2.87 moles of sodium chloride (Nacl) into a clean dry cup. ​
Anettt [7]

Answer: weigh is m = n × M = 2.87 mol × 58.44 g/mol

Explanation: mass = amount of substance × molar mass

M((NaCl) = 22.99 +35.45

6 0
3 years ago
A gas has a volume of 340.0 mL at 45.90 oC. What is the new temperature of the gas, in Kelvin, if the volume increased to 550.0
d1i1m1o1n [39]

Answer:

Solution for A gas has a volume of 340.0 mL at 45.90 degree celsius. What is the new temperature of the gas, in kelvin, if the volume increased to 550.0 mL.

Missing: oC. ‎| Must include: oC.

Explanation:

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soldier1979 [14.2K]
HCL 2.0x 3.6790= KOH SOLITION
6 0
3 years ago
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