Answer:
204g of NH3
Explanation:
The balanced equation for the reaction is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the number of mole NH3 produced by reacting 6moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 6 moles of N2 will react to produce = 6 x 2 = 12 moles of NH3.
Finally, we shall convert 12 moles of NH3 to grams. This is illustrated below:
Number of mole of NH3 = 12 moles.
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 =..?
Mass = mole x molar mass
Mass of NH3 = 12 x 17
Mass of NH3 = 204g.
Therefore, 204g of NH3 will be produced from the reaction.
The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is <u>3.347</u>.
Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.
<u>Calculation:-</u>
Normality of acid Normality of base
= nMV nMV
= 1 × 0. 15 × 0.017 1 × 0. 20 ×0.015 L
= 2.55 × 10⁻³ = 3 × 10⁻³
The overall base will be high
net concentration = 3× 10⁻³ - 2.55 × 10⁻³
= 0.45 × 10⁻³
= 4.5× 10⁻⁴
pH = -log[4.5 × 10⁻⁴]
= 4 - log4.4
= <u>3.347</u>
A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.
Learn more about titration here:-brainly.com/question/186765
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Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.
Answer:
0.456 M
Explanation:
Step 1: Write the balanced neutralization equation
HNO₂ + KOH ⇒ KNO₂ + H₂O
Step 2: Calculate the reacting moles of KOH
9.26 mL of 1.235 M KOH react.
0.00926 L × 1.235 mol/L = 0.0114 mol
Step 3: Calculate the reacting moles of HNO₂
The molar ratio of HNO₂ to KOH is 1:1. The reacting moles of HNO₂ are 1/1 × 0.0114 mol = 0.0114 mol.
Step 4: Calculate the initial concentration of HNO₂
0.0114 moles of HNO₂ are in 25.0 mL of solution.
[HNO₂] = 0.0114 mol / 0.0250 L = 0.456 M
Hello there.
<span>The rate of evaporation is affected by all of the following except
Answer: </span><span>B. the humidity.
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