Answer: No
Explanation: For it to be a divergent boundary, the arrows would have to be pointing in opposite directions. (one points left, one points right).
Answer:
Just click on the answer, if it is an external link DO NOT CLICK IT. it may contain dangerous stuff including h*cked stuff.
Explanation:
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles
n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?
n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles
Agt STP 1 mole= 22.4 L what obout 0.178 moles
= 22.4 x0.178moles/ 1moles =3.98 L( answer C)