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3 years ago

The incomplete table below shows selected characteristics of gas laws.

Chemistry

2 answers:

lorasvet [3.4K]3 years ago

6 0

Answer:

pressure and temperature

Explanation:

liq [111]3 years ago

4 0

Answer:

Pressure and temperature.

Explanation:

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What has a higher density Ice cubes or ice spheres pls pls pls help

Verizon [17]

The answer is Ice spheres

6 0

2 years ago

⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

3 years ago

What is energy in motion called

oee [108]

Is it kinetic energy

3 0

3 years ago

Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of

jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = $2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x$

$2330.35x=1048815\\x=450.068g$

Hence, there was 450.068g of water in the pot.

8 0

2 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

2 years ago