Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%
m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water
%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
Answer:
If the nitrogen atom is a neutral atom, it will have seven electrons orbiting the nucleus of the atom. This is because neutral atoms get their neutral...
Explanation:
Answer:
Explanation:
Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.
Multiply both sides of the equation by 6.8 moles of krypton.
The units of moles of krypton will cancel.
The denominator of 1 can be ignored, so this becomes a simple multiplication problem.
If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.
6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.
The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.576 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.019862 mol
(101,183.9 Pa)V = (0.019862 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L
