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Blizzard [7]
2 years ago
12

The photo shows a frozen juice pop on a hot day.

Chemistry
2 answers:
GaryK [48]2 years ago
6 0

Answer:

Arrows pointing from the sun to the juice pop

Explanation: Trust I got it right

Alekssandra [29.7K]2 years ago
3 0

Answer:

C! Hope this helps !:)

Explanation:

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What particle is J.J Thomson credited with discovering
Ilya [14]
He was credited with discovering the subatomic particle also known as the electron in 1897.
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3 years ago
In a titration, 10.0 ml of 0.0750 M HCl(aq) is exactly neutralized by 30.0 ml of KOH(aq) of unknown concentration. What is the c
Sphinxa [80]

Answer: 0.0250

Explanation: 10 X 0.0750 = .75

.75 / 30 = 0.0250 M

4 0
3 years ago
A solution that holds more solute than it normally can under the conditions at a given temperature is _____.
Setler79 [48]
It would be super saturated
3 0
3 years ago
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What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?
kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of Al(OH)_3 = 0.610 g

Molar mass of Al(OH)_3 = 78 g/mol

\text{Number of moles}=\frac{0.610g}{78g/mol}

Number of moles of Al(OH)_3 = 0.0078 moles

The reaction between Al(OH)_3 and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

By Stoichiometry,

1 mole of  Al(OH)_3 reacts with 3 moles of HCl

So, 0.0078 moles of Al(OH)_3 will react with \frac{3}{1}\times 0.0078 = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

0.0234moles=\frac{\text{Given mass}}{36.5g/mol}

Mass of HCl required will be = 0.8541 grams

3 0
3 years ago
HELP ASAP PLEASE!!!! 30 POINTS
valentina_108 [34]

Answer:

it's option c

Explanation:

because if I'm not wrong I have learned these type of questions back 11 and I remember that rutherfords observation was few alpha particles were deflected by small angles.

5 0
3 years ago
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