Answer:
The mass ratio of zinc to sulfide is 85:42.
2.5559 kg of Zn are in 3.82 kg of ZnS.
Explanation:
a) Mass of zinc sulfide = 254 g
Mass of zinc in a zinc sulfide sample = 170 g
Mass of sulfide in zinc sulfide sample = x
254 g = 170 g+ x
x = 84 g
The mass ratio of zinc to sulfide:
b) Mass of zincsulfide sample = 3.83 kg
The mass ratio of zinc to sulfide is 85:42.
Let the mass of zinc and sulfide be 85x and 42x respectively:
85 x+ 42 x=3.82 kg
x =0.03007 kg
Mass of an zinc= 85x=85 × 0.03007 kg= 2.5559 kg
Converting temperature of 68°F to °C gives 20 °C.
Converting temperature of 68°F to K gives 293 K.
<h3>What is temperature conversion?</h3>
Temperature conversion is the process of converting the measurement units of the temperature recorded in a particular unit to another unit.
The various units of Temperature include;
- degree Celsius
- degree Fahrenheit
- degree Kelvin
Temperature is measured with thermometer and it records the hotness or coldness of a body.
<h3>Converting 68°F to °C</h3>
F = 1.8C + 32
(F - 32/1.8) = C
(68 - 32) / 1.8 = C
20 ⁰C = 68 ⁰F
<h3>Converting 20°C to K</h3>
0 °C = 273 K
20 °C = 273 + 20 = 293 K
Learn more about temperature conversion here: brainly.com/question/23419049
#SPJ1
Answer:
F=ma
Explanation:
F=m×a
according to that F÷m=a and also F ÷a=m
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
