The partial pressure of oxygen in a sample of air increases if the temperature is increased.
Answer: Option 1
<u>Explanation:
</u>
According to Guy-Lussac's law, at constant volume, pressure exhibited by the gas molecules will be directly proportional to the temperature of the gas molecules. It is also known that pressure of mixture of gas molecules is the sum of partial pressure of each gas molecule in the mixture.
If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.
So, it can also be concluded that temperature increases the oxygen's partial pressure in air increases.
Answer:
dollars each person will receive.
Explanation:
Number of people in which 1 mole of pennies is distributed = 250 million =

250 million =
persons
Number of pennies in 1 mole = 
Pennies per person:

1 penny = 0.01 $

dollars each person will receive.
Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
Molarity = moles of solute(HCl)
------------------------------------
volume of the solution
= 1
------
5
= 0.2M.
Hence option B is correct.
Hope this helps!!
Answer:
Heat given off was -34.34kJ
Explanation:
Mass of iron bar = 869g
Initial temperature (T1) = 94°C
Final temperature (T2) = 5°C
Specific heat capacity of iron (c) = 0.444J/g°C
Heat energy (Q) = Mc∇T
Q = heat energy
c = specific heat capacity
∇T = change in temperature
M = mass of the substance
Q = mc∇T
∇T = T2 - T1
Q = Mc(T2 -T1)
Q = 869 * 0.444 * (5 - 94)
Q = 385.836 * -89
Q = -34339.404J
Q = -34.34kJ
The heat given of was -34.34kJ