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2 years ago

Cool air over warm water will:

Chemistry

1 answer:

navik [9.2K]2 years ago

3 0

Answer:

Air and Water Temperature Increases

An increase in the air temperature will cause water temperatures to increase as well. ... Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.

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At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

3 years ago

What happens when acetylene reacts with silver powder?

horrorfan [7]

It dissolves I think I know I am expert and. Ute

6 0

2 years ago

How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti

kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is $58.4\ g/mole$ .

To Find :

How many grams of NaCl is required .

Solution :

Let , NaCl required is x gram .

Molarity is given by :

$M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g$

Hence , this is the required solution.

6 0

3 years ago

If i have an 86 and i don’t do a 50 point assignment, what will my grade be in the class?

nexus9112 [7]

Well you are at a 86% so t<span>his is considered a "B" grade on an average grade scale. If you take 50 points we would need to know the Total point you could have got in that class to be able to see how much a percent was the assignment work and then take that percentage of the assignment and subtract it from the 86% to see where on the scale you would fall in after the assignment points were taken off. Hope this helps :)
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7 0

3 years ago

How many moles of propane<br> react when 294 g of CO2 form?<br><br> C3H8 +502 → 3CO₂ + 4H₂O

tatiyna

2.23 moles of propane react when 294 g of CO₂ is formed .

<h3>What is moles ?</h3>

Moles is a unit which is equal to the molar mass of an element.

A reaction is given

C₃H₈ +50₂ → 3CO₂ + 4H₂O

Grams of CO₂ formed = 294 gm

In moles = 294 /44 = 6.68 moles.

Let x be the moles of C₃H₈ is x

Mole ratio of CO₂ to C₃H₈ = 3 : 1

6.68 /x = 3/1

x = 6.68 /3 = 2.23 moles

Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .

To know more about Moles

brainly.com/question/26416088

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7 0

2 years ago