Answer:
The moles of ammonia present in a 1.284 g sample are 0.075.
Explanation:
Molar mass is the mass of one mole of a substance, which can be an element or a compound. If the molar mass of ammonia is 17,030 g / mol, then in 1 mole of ammonia there are 17,030 g.
So, in this case, the following rule of three can be applied: if by definition of molar mass in there are 17.030 g in 1 mole of ammonia, 1.284 g of ammonia in how many moles will it be?
moles=0.075
<u><em>The moles of ammonia present in a 1.284 g sample are 0.075.</em></u>
Experiment
When making a guess and retesting this information, a model or <u>experiment</u> may be formed which explains why something has occured or what it may look like.
When a solid turns to gas it is called sublimation, and when a gas turns into a liquid it is called deposition
<span>1. </span>To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
V2 = 104.1 x 478 / 88.2
<span> V2 =564.17 cm^3</span>
144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.
<h3>What is Ideal Gas Law ? </h3>
The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.
PV = nRT
where,
P = Presure
V = Volume in liters
n = number of moles of gas
R = Ideal gas constant
T = temperature in Kelvin
Here,
P = 1 atm [At STP]
R = 0.0821 atm.L/mol.K
T = 273 K [At STP]
Now first find the number of moles
F₂ + CaBr₂ → CaF₂ + Br₂
Here 1 mole of F₂ reacts with 1 mole of CaBr₂.
So, 199.89 g CaBr₂ reacts with = 1 mole of F₂
1.28 g of CaBr₂ will react with = n mole of F₂
n = 0.0064 mole
Now put the value in above equation we get
PV = nRT
1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K
V = 0.1434 L
V ≈ 144 mL
Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.
Learn more about the Ideal Gas here: brainly.com/question/20348074
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