Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
Answer: Sodium bromide is an ionically bonded compound.
(NaBr: Sodium Bromide)
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
Answer:
c) 2.5 mL
Explanation:
Solution
Doctors order = 0.125g
and
The liquid suspension concentration = 250 mg/5ml
= 0.250g/5ml
Or 0.05g/ml
Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml
