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3 years ago

Question 18 of 35

Chemistry

1 answer:

sammy [17]3 years ago

4 0

Answer:

A. Dark-colored moths become common after tree trunks become

dark.

Explanation:

I took the test

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Name four products of incomplete combustion 1 - ____ 2 -____ 3 -____ 4 -_____

faltersainse [42]

Answer:

Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide

Explanation:

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3 years ago

What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?

blagie [28]

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= ?
n = 0.5 moleT= 273.15 K (at STP)
p= 101.325 kPa (at STP)
R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:
V=nxRxT/p
V=0.5x8.314x273.15/101.325= 11.2 L = 11200 mL

Answer: D.

5 0

3 years ago

A solution is produced in which water is the solvent and there are four solutes. Which of the solutes can dissolve better if the

djverab [1.8K]

Answer:

What are the solutes

Explanation:

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4 years ago

Which organelle if empty would cause the plant to wilt

valentina_108 [34]

The vacuole because the less water it has the more it will wilt. The reason of it wilting is because the cells shrink which is the cause of wilting.

8 0

4 years ago

How do I do this question?The aluminum cup inside your calorimeter weighs 39.78 g. You add 50.01 g of ice cold water to the calo

konstantin123 [22]

Answer:

$Cp_{metal}=0.922\frac{J}{g\°C}$

Explanation:

Hello.

In this problem we must realize that there is heat flow that moves from the hot metal object and the hot water to the cold water and the cold aluminum cup, which allows us to write:

$Q_{cup}+Q_{cold,w}=-(Q_{metal}+Q_{hot,w})$

Which means that the heat lost be the hot metal object and the hot water is gained by both the cold water and the cold aluminum cup, which can be written in terms of mass, specific heats and change in temperature towards the equilibrium temperature (35.9 °C):

$m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})=-(m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})$

We need to solve for the specific heat of the metal as shown below:

$Cp_{metal}=\frac{m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})}{-m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{39.78g*0.903\frac{J}{g\°C}(35.9-0.5)\°C+50.01g*4.184\frac{J}{g\°C}(35.9-0.5)\°C +50.72g*4.184\frac{J}{g\°C}(35.9-69.5)\°C }{-49.98g(35.9-69.5)\°C } \\\\Cp_{metal}=\frac{1271.6J+7407.2J-7130.3J}{-1679.3g\°C} \\\\Cp_{metal}=0.922\frac{J}{g\°C}$ Best regards.

3 0

3 years ago