The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
Answer:
Ne
Explanation:
Atomic number of Ne is 10.
Electronic configuration of Ne:

Octet of Ne is complete . Element having complete octet are stable and behave ideal gas.
and
are reactive and hence, does not behave as ideal gas.
Layla it is A. that's the only one in standard form.
Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,


345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
Answer:
110000
Explanation:
Multiply 2.75 by 4. is 11
Multiply 11 by 10000. is
110000