Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:
Number of moles (n)
can be written as: 
where, m = given mass
M = molar mass
where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant = 
T = temperature of the gas = 
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Answer:
CaCO3 is false
Explanation:
Because HCl is hydrongen chloride
Answer:
Each molecule contains one atom of A and one atom of B. The reaction does not use all of the atoms to form compounds.
A + B ⟶ Product
Particles: 6 8 6
If six A atoms form six product molecules, each molecule can contain only one A atom.
The formula of the product is ABₙ.
If n = 1, we need six atoms of B.
If n = 2, we need 12 atoms of B. However, we have only eight atoms of B, so the formula of the product must be AB.
Thus, 6A + 6B ⟶ 6AB, with two B atoms left over.
Explanation:
Credit goes to @znk
Hope it helps you :))
The equation is
W = C/F
W= 3.00 x 10^8 m/sec
——————————
6.165 x 10^14 Hz
W= 4.87 x 10^-7 m
Energy is
E=hF
E= (6.626 x 10^-34 Jxsec )(6.165 x 10^14 Hz)
E= 4.085 x 10^-19 J
Answer:
3.0 moles Al₂O₃
Explanation:
We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.
4 Al + 3 O₂ -----> 2 Al₂O₃
6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al
4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂
As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.
However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.
