Answer:
29.42 Litres
Explanation:
The general/ideal gas equation is used to solve this question as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K
According to the information provided in this question;
mass of nitrogen gas (N2) = 25g
Pressure = 0.785 atm
Temperature = 315K
Volume = ?
To calculate the number of moles (n) of N2, we use:
mole = mass/molar mass
Molar mass of N2 = 14(2) = 28g/mol
mole = 25/28
mole = 0.893mol
Using PV = nRT
V = nRT/P
V = (0.893 × 0.0821 × 315) ÷ 0.785
V = 23.09 ÷ 0.785
V = 29.42 Litres
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
--------------------
Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
--------------------
Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**
Answer:
1 mole represents 6.023×1023 particles.
1 mole of iodine atom= 6.023×1023
Given 127.0g of iodine.
no. of iodine atom = 1 mole of iodine
1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg
Given 48g of Mg = 2×6.023×1023
no. of Mg = 2 moles of Mg
1 mole of chlorine atom= 6.023× 1023
no. of chlorine atom = 35.5g of chlorine atom
Given 71g of chlorine atom=2× 6.023× 1023
no. of chlorine atom = 6.023×1023
2 moles of chlorine atom.
Given that 4g of hydrogen atom.
will be equal to 4 × 6.023 × 1023
no. of atoms of hydrogen= 4 moles of hydrogen atom.
I think that you should use your phone
