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kodGreya [7K]
2 years ago
11

If a pH of water is 5.3 then 4.3 is it 1.0 less acidic 1.0 less basic

Chemistry
2 answers:
Sedbober [7]2 years ago
7 0
It would be 1.0 less basic, 1.0 more acidic
Gnom [1K]2 years ago
4 0
The pH of water is 7
5.3 to 4.3 is more acidic. 1 pH is more acidic than 2 pH The base starts at 8.
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Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g
Alex787 [66]
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
8 0
3 years ago
Read 2 more answers
What happens when you put magnesium carbonate and diluted water
Rina8888 [55]

Answer:

Magnesium carbonate doesn't dissolve in water, only acid, where it will effervesce (bubble).

Explanation:

 An insoluble metal carbonate reacts with a dilute acid to form a soluble salt. Magnesium carbonate, a white solid, and dilute sulfuric acid react to produce magnesium sulfate. Colourless magnesium sulfate heptahydrate crystals are obtained from this solution.

8 0
3 years ago
Write the trigonometric equation for the function with a period of 6. The function has a maximum of 3 at x = 2 and a low point o
White raven [17]

Answer:

Do you have a picture?

Explanation:

Sometimes, pictures are required to answer questions.

7 0
3 years ago
Chlorine atoms react with methane, forming HCl and CH3. The rate constant for the reaction was determined to be 3.600×107 at 278
Ratling [72]

<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:

\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{317K} = equilibrium constant at 317 K = 3.050\times 10^{8}M^{-1}s^{-1}

K_{278K} = equilibrium constant at 278 K = 3.600\times 10^{7}M^{-1}s^{-1}

E_a = Activation energy = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 278 K

T_2 = final temperature = 317 K

Putting values in above equation, we get:

\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol

Hence, the activation energy for the reaction is 40.143 kJ/mol

8 0
3 years ago
A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture
Elenna [48]

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

8 0
3 years ago
Read 2 more answers
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