Determine the density of CO2 gas at STP

The total volume of seawater is 1.5 x 1021L .Assume that seawater contains 3.1 percent sodium chloride by mass and that its dens

son4ous [18]

Answer:

Mass in kg = 4.7*10^19 kg

Mass in tons = 5.2*10^16 tons

Explanation:

Given:

Total volume of sea water = 1.5*10^21 L

Mass % NaCl in seawater = 3.1%

Density of seawater = 1.03 g/ml

To determine:

Total mass of NaCl in kg and in tons

Calculation:

Unit conversion:

1 L = 1000 ml

The volume of seawater in ml is:

$=\frac{1.5*10^{21}L*1000ml }{1L} =1.5*10^{24} ml$

$Mass\ seawater = Density*volume = 1.03g/ml*1.5*10^{24} ml=1.5*10^{24}g$

$Mass\ NaCl\ in\ seawater = \frac{3.1}{100}*1.5*10^{24} g=4.7*10^{22} g$

To convert mass from g to Kg:

1000 g = 1 kg

$Mass\ seawater(kg) = \frac{4.7*10^{22}g*1kg }{1000g} =4.7*10^{19} kg$

To convert mass from g to tons:

1 ton = 9.072*10^6 g

$Mass\ seawater(tons) = \frac{4.7*10^{22}g*1ton }{9.072*10^{6}g } =5.2*10^{16} tons$

5 0

3 years ago

A gas with a volume of 4.0L at a pressure of 205 kPa is allowed to expand to a volume of 12000 mL. What is the pressure in atmos

Leni [432]

Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 205 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 4.0 L

$V_2$ = final volume of gas = 12000 ml = 12 L (1L=1000ml)

$205\times 4.0=P_2\times 12$

$P_2=68.3kPa=0.674atm$ (1kPa=0.0098atm)

Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.

8 0

3 years ago

What is the mass of 3.62X10^24 molecules of methanol

Bad White [126]

N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.

7 0

3 years ago

Read 2 more answers

A dry mixture of KNO3 and sand could be separated by

Tatiana [17]

Place the mixture in hot water and stir well.
The KNO3 is very soluble in hot water. 
Use a fine filter paper and filter off the sand. 
The sand will be separated from the KNO3 solution. 
The water can now be evaporated from the solution by further, gentle heating leaving the solid in the container.

7 0

3 years ago

A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to

12345 [234]

Answer:

The resulting solution contains approximately 666 g of water.

Explanation:

In the initial solution we have:

1g salt : 8g sugar : 200g water

This means that the ratios are:

$\frac{salt}{sugar} = \frac{1}{8} \\\\\frac{sugar}{water} = \frac{8}{200} =\frac{1}{25}$