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3 years ago

Can someone please help me with this?

Chemistry

2 answers:

gladu [14]3 years ago

4 0

Answer:

Ball 2

Explanation:

the ball is falling and when it hits the ground it will bounce back up creating potential enery

Yuliya22 [10]3 years ago

4 0

Answer:

I think It's ball 1

Explanation:

I could be wrong sorry if I am

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3 years ago

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Which is not a physical property of water? weak adhesion with glass strong cohesion strong surface tension ability to move up th

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3 years ago

How many moles of H2SO4 are present in 3.63 liters of a 0.954 M solution?

azamat

For this problem, we will use the formula setup for Molarity.
The formula is M= moles
---------
liters

So, lets plug in the numbers we have. We know M= 0.954, and L (liters)= 3.63. Lets plug that into our equation.

x
0.954= -----
3.63

As we can see, our moles are unknown.
So, lets multiply 0.954 times 3.63 to get our moles.
0.954 x 3.63 equals 3.46302.

We know this is our correct answer, because when we divide 3.46302 and 3.63, we get 0.954.

Why?
Because when we divide the moles by liters, we get the molarity.

I hope I helped!!!!!

Here are some common units to know if my units confused you.
M= Molarity, which is what we find by dividing moles by liters
mol= Moles
L= Liters

5 0

4 years ago

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 li

Lera25 [3.4K]

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

$[CO]_i = \frac{0.500mol}{5.00L} = 0.100M$

$[H_2]_i = \frac{1.500mol}{5.00L} = 0.300M$

$[H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M$

Step 3: Make an ICE chart

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I 0.100 0.300 0 0

C -x -3x +x +x

E 0.100-x 0.300-3x x x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

$Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41$

3 0

4 years ago