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kolbaska11 [484]
2 years ago
14

PLEASE HELP!

Chemistry
1 answer:
Tanzania [10]2 years ago
3 0
Molar mass:

KCl = 74.55 g/mol

KClO3 = 122. 55 g/mol

  <span>Calculation of the mass of KClO3 :</span>

<span>2 KClO3 =  2 KCl + 3 O2</span>

2* 122.55 g KClO3 ------------------ 2 * 74.55 g KCl
mass KClO3 ?? --------------------- 25.6 g KCl

mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55

mass KClO3 = 6274.56 / 149.1

mass = 42.082 g of  KClO3

Therefore:

1 mole KClO3 ---------------------- 122.55 g
?? moles KClO3 ------------------- 42.082 g

moles KClO3 = 42.082 * 1 / 122.55

moles KClO3 = 42.082 / 122.55

=> 0.343 moles of KClO3


Answer C

hope this helps!

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covalent

Explanation:

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8 0
3 years ago
A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is
Romashka [77]

Answer:

\large \boxed{\text{B.) 2.8 atm}}

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\

(c) Convert the pressure to atmospheres

p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}

7 0
2 years ago
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
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Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

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pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

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Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

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4 0
3 years ago
How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
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4% mass / volume :

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V = 1.2 * 100 / 4

V = 120 / 4

V = 30 mL

hope this helps!

7 0
2 years ago
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