Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
Is there an equation? I can't help if there's no equation involved.
1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
It is calculated in mg/ml.
The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.
Formula for calculating concentration in mg/ml is
Conc. (mg/ml) = M(eq) /ml × Molecular weight / Valency
Given
M(eq) NaCl/ ml = 23.5
Molecular weight pf NaCl = 58.5 g/mol
Valency = 1
Putting the values into the formula
Conc. (mg/ml) = 23.5 ×58.5/1
= 1374.75 mg/ml
Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
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Answer:
The value of the heat capacity of the Calorimeter 
= 54.4 
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water 
= 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q = 
ΔT + ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 + × 11.24
611.37 = × 11.24
= 54.4
This is the value of the heat capacity of the Calorimeter.
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
