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3 years ago

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Balance the following chemical equation

1 answer:

babymother [125]3 years ago

6 0

Hello:) The answers are in the picture below. I’m not sure how to explain but hope it helps :)

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V1 = l; P1=40Pa; P2= 100 kpa; V2= 1.0 L please help!

blsea [12.9K]

Answer:

V1 = 2500 L

Explanation:

V1P1 = V2P2

x * 40 = 100.000 * 1.0

==> x =2500 L

6 0

3 years ago

A gas occupies a volume of 650.0 mL when the pressure is 3.50 atm. What will the new volume be if the pressure is reduced to 1.6

Gemiola [76]

The new volume will be 1379 mL.

Explanation:

As per Boyle's law, the product of initial volume and initial pressure of any gas molecule is equal to the product of final volume and final pressure of those molecules.

So here the initial volume is 650 ml and the initial pressure is 3.50 atm. As the temperature is said to be constant, then this system will be obeying Boyle's law. So, the final pressure is given as 1.65 atm. As there is a reduction in the pressure, the volume of the gas is tend to get expanded.

$P_{1} V_{1} = P_{2} V_{2}$

So, $(650*10^{-3}*3.50)=(1.65*V_{2})$

$V_{2} = \frac{650*10^{-3}*3.50}{1.65} = 1379 mL$

So, the new volume of the gas on reduction in pressure is 1379 mL.

7 0

3 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$0.173\%=\frac{779 mg}{m}\times 100$

$m=\frac{779 mg}{0.173}\times 100= 450,289 mg$

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0

3 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago

List the protons, neutrons, and electrons for Ge2+<br> Please hurry

Nady [450]

Answer:

32, 30 and 41

Explanation:

The problem here is to find the number of:

Protons, neutrons and electrons in Ge²⁺

In this ion,

We must understand that for a net positive charge to remain on an atom, the number of protons must be greater than the number of electrons.

Ge is Germanium with atomic number of 32;

So the number of protons is 32

Since the atom has lost two electrons;

Number of electrons now is 32 - 2 = 30

Number of neutrons is 41 from the periodic table.

7 0

2 years ago