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2 years ago

9

Can someone answer these questions about the LACK OF CLEAN DRINKING WATER? PLS. I will give 30 pts.

1 answer:

Andru [333]2 years ago

8 0

Answer:

Answer

Explanation:

Zack Gomez

Lack of clean water

Water should be clean cause we need water so we can drink it cause we live off of it so we need it so it should be clean. Animals need it to survive in the hot so they need it so it should be clean. Insects need it to survive so they need it so it should be clean. THE END

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Write an ionic equation of hydrogen peroxide reacting with sodium sulphite

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Answer:

Na2SO3 + H2O2 = Na2SO4 + H2O

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A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new

dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP = $10^5Pa$

$P_2$ = final pressure of gas = ?

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$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = $30^oC=273+30=303K$

Now put all the given values in the above equation, we get:

$\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}$

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Read 2 more answers

What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of

Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 = -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m= molality

molality = $\frac{moles of solute}{mass of solvent in kilograms}$

putting the value in the equation:

molality= $\frac{1.31}{0.53}$

= 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

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