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2 years ago

A ____________ is a large area of open land that has been turned into a container for our trash.

Chemistry

2 answers:

sdas [7]2 years ago

4 0

The correct answer is Landfill
:)

gregori [183]2 years ago

4 0

Answer:

Landfill (D) is your answer to go in the blank.

Explanation:

The garbage truck is what transports our garbage and delivers it to the landfill.

Housefill isn't a thing.

Recycling Bin, is how you recycle and goes to a facility where it is modified to be re-used, and isn't transported to a landfill.

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At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressur

fenix001 [56]

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the partial pressure of He would give a solubility of 0.200 M.

<h3>What is Henry's Law ?</h3>

Henry's Law is a gas law states that at a constant temperature the amount of gas that dissolved in a liquid is directly proportional to the partial pressure of that gas.

<h3>What is relationship between Henry's Law constant and Solubility ?</h3>

The solubility of gas is directly proportional to partial pressure.

It is expressed as:

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

where,

$S_{\text{gas}}$ = Solubility of gas

$K_{H}$ = Henry's Law constant

$P_{\text{gas}}$ = Partial pressure of gas

Now put the values in above expression we get

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

0.080M = $K_{H}$ × 1.7 atm

$K_{H} = \frac{0.080\ M}{1.7\ \text{atm}}$

= 0.047 M/atm

Now we have to find the partial pressure of He

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

0.200 M = 0.047 M/atm × $P_{\text{gas}}$

$P_{\text{gas}} = \frac{0.200 M}{0.047\ \text{M/atm}}$

= 4.25 atm

Thus from the above conclusion we can say that At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the partial pressure of He would give a solubility of 0.200 M.

Learn more about the Henry's Law here: brainly.com/question/23204201

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3 0

1 year ago

Explain intermolecular forces cause water molecules to stick to a material like a glass pitcher

natulia [17]

Answer:

Cohesive forces cause the water molecules to stick together with a lot of elasticity, allowing the water to function very much like a rubber into a material like a glass pitcher.

6 0

2 years ago

What is the mass in grams of ba(io3)2 can be dissolved in 500 ml of water at 25 degrees celcius?

lesya [120]

The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g

<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>

The Ksp of Ba(IO3)2 = 1.57 × 10^-9

Molar mass of Ba(IO3)2 = 487 g/mol?

Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

$Ba(IO_{3})_{2} \leftrightharpoons Ba^{2+} + 2\:IO_{3}^{-}$

$Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}$

$[Ba(IO_{3})_{2}] = \sqrt[3]{ksp} =\sqrt[3]{1.57 \times {10}^{ - 9} } \\ [Ba(IO_{3})_{2}] = 1.16 \times {10}^{-3} moldm^{-3}$

moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles

mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g

Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.

Learn more about mass and moles at: brainly.com/question/15374113

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7 0

1 year ago

Write the overall reaction for cellular respiration

goblinko [34]

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

7 0

2 years ago

A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago