Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Explanation:
Pressure in the submarine = 108.9 kPa
Volume, V = 2.4 * 10^5 L
Pressure, P = 116k Pa
Temperature, T = 312 K
Ideal gas law: PV = nRT or n = PV / RT
So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K
= 1.073 *10^4 mol
when temperature is changed to 293K,
PV = nRT or P = nRT / V
=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L
=108.9 K Pa
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
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Answer:
The main purpose for converting numbers into scientific notation is to make calculations with surpisingly large or small numbers less complicated. Since zeros are not used to place the decimal point, all of the digits in a number in scientific notation are meaningful and easier to read. Hope this helps!
Explanation:
Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.