A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask<span>The given are: </span>
<span><span>1. </span>Mass = ?</span><span><span /></span>
<span><span>2. </span>Density = 1298 g/L</span>
3. Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L
<span>Formula and derivation: </span><span><span>
1. </span>density = mass / volume</span> <span><span>
2. mass </span>= density / volume</span>
<span>Solution for the problem: </span><span><span>
1. mass = </span></span> <span> 1298 g/L / 0.125 L = 10384g
</span>
Answer:
Explanation:
3.
Knowns: 100mL of solution; concentration of 0.7M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 100/1000 * 0.7 = 0.07 mole
Final Answer: 0.07mole
2.
Knowns: 5.50L of solution; concentration of 0.400M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 5.5 * 0.4 = 2.20 mole
Final Answer: 2.20 mole
In order to solve this, we need to make use of Hess' Law.
We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)
Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)]
</span>
The person above is trying to give you a virus
Answer:
b)15.0°C
Explanation:
Specific Heat of Water=4.2 J/g°C
This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.
∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.
80×4.2 J=336 J
Total Energy Provided=1680 J
The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}
Temperature increase=
=5°C
Initial Temperature =10°C
Final Temperature=Initial + Increase in Temperature
=10+5=15°C
