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Sunny_sXe [5.5K]
2 years ago
12

2 points

Chemistry
1 answer:
Taya2010 [7]2 years ago
7 0
Single Replacement bc Al replaces H and displaces H making it hydrogen gas H2
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Hi does anyone have the data table answers for the Charles Law Lab Report from Edge?
ZanzabumX [31]

Answer:

Explanation:

hope its not too late :)

6 0
3 years ago
A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16
Alja [10]

the mass of ice taken = 10 g


the mass of water = 250 g


initial temperature of water = 20 C


the final temperature of water = 16. 8 C


specific heat of water = 4.18 J/g*K


the heat absorbed by ice to melt = heat loss by water


heat loss by water = mass X specific heat of water X change in temperature


heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules


heat gained by ice = 3344 J


heat gained by ice = enthalpy of fusion X moles of ice


moles of ice = mass / molar mass = 10 / 18 = 0.56 moles


enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole


3 0
3 years ago
The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). If an object’s mas
Amiraneli [1.4K]
Weight  = mass x gravitational acceleration = mg

m = 10kg
g = 9.8 m/s^2

So, weight =  10 x 9.8  = 98 Kg-m/s^2
 


5 0
3 years ago
Read 2 more answers
Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0
3 years ago
You
Igoryamba

Answer:

The arm that was not sprayed with anything

Explanation:

The control group would be <u>the arm that was not sprayed with anything</u>.

<em>The control group during an experiment is a group that forms the baseline for comparison in other to determine the effects of a treatment. The control group does not include the variable that is being tested and as such, it provides the benchmark to measure the effects of the tested variable on the other group - the experimental group. In this case, the experimental group would be the arm that was sprayed with the repellent.</em>

8 0
3 years ago
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