Compare and contrast the different way heat moves

when fully inflated a hot air balloon has a volume of 1.6 *10^5 L ( liter) an average temperature of 373k and 0 967 atm . 1) Ass

Sidana [21]

1. First, you have to find the number of moles 1.6z10^5L of gas is at 373K and 0.967atm using PV=nRT solving for n. (n=PV/RT). Everything is in the correct units and we know R is going to be 0.08206atmL/molK since it is a constant.
n=(0.967atmx160000L)/(0.08206atmL/molKx373K)
n=5054.8mol gas
Then you have to find the the number grams which can be found using the molar mass given as 29g/mol. multiply 29g/mol by the number of moles of gas we found in the previous step.
5054.8molx29g/mol=146589.9g of gas
Lastly, to find the density of the gas you need to divide the mass of the gas by its volume.
146589.9g/160000L=0.916g/L

2. The dinsity of the gas at STP should be higher than the density of gas with the given conditions. This is due to the fact that the given conditions involves a higher temperature than that of at STP which will cause the gas to expand therefore increasing the volume with out increasing the mass. The reason why the pressure is not building up even though the pressure is higher is that the balloon is not sealed meaning the gas can maintain about atmospheric pressure while expanding since the excess are just leaves the balloon.
the answer to part 2 can be proven by the fallowing:
To find the density of the gas at STP you first multiply the molar volume of gas at STP by the number of moles of gas from part 1 to get the volume of the gas at STP.
5054.8molx22.4L/mol=113228L
Then you divide the mass form part by the new volume to get the new density.
146589.9g/113228L=1.30g/L

I hope this helps. Let me know in the comments if any of it is unclear.

4 0

3 years ago

Read 2 more answers

Be sure to answer all parts.

tangare [24]

Answer:

(a) cesium bromide (CsBr): 9.15 grams

(b) calcium sulfate (CaSO4): 5.85 grams

(c) sodium phosphate (Na3PO4): 7.05 grams

(d) lithium dichromate (Li2Cr2O7): 9.88 grams

(e) potassium oxalate (K2C2O4): 7.15 grams

Explanation:

(a) cesium bromide (CsBr):

Molar mass of CsBR = 212.81 g/mol

Number of moles = molarity * volume

Number of moles = 0.100 M *0.43 L

Number of moles = 0.043 moles

Mass of CsBr required = moles * Molar mass

Mass of CsBr required = 0.043 moles * 212.81 g/mol

Mass of CsBr required = 9.15 grams

(b) calcium sulfate (CaSO4):

Molar mass of CaSO4 = 136.14 g/mol

Mass of CaSO4 required = moles * Molar mass

Mass of CaSO4 required = 0.043 moles * 136.14 g/mol

Mass of CaSO4 required = 5.85 grams

(c) sodium phosphate (Na3PO4):

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 required = moles * Molar mass

Mass of Na3PO4 required = 0.043 moles * 163.94 g/mol

Mass of Na3PO4 required = 7.05 grams

(d) lithium dichromate (Li2Cr2O7):

Molar mass of Li2Cr2O7 = 229.87 g/mol

Mass of Li2Cr2O7 required = moles * Molar mass

Mass of Li2Cr2O7 required = 0.043 moles * 229.87 g/mol

Mass of Li2Cr2O7 required = 9.88 grams

(e) potassium oxalate (K2C2O4):

Molar mass of K2C2O4 = 166.22 g/mol

Mass of K2C2O4 required = moles * Molar mass

Mass of K2C2O4 required = 0.043 moles * 166.22 g/mol

Mass of K2C2O4 required = 7.15 grams

7 0

2 years ago

If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.

kati45 [8]

This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

$M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M$

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

$pH=-log(0.0333)\\\\pH=1.477$