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svetlana [45]
3 years ago
15

First answer will be brainliest!!

Chemistry
1 answer:
igomit [66]3 years ago
8 0
I think it is rarefaction. But im not sure

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Help. The answer to question 1 I didn't mean to click so please answer that as well.
Zepler [3.9K]
We have that energy=specific heat * change in temperature * mass. Thus, we have the final temperature (22) minus the initial temperature (55) to equal -33 as our change in temperature. Our specific heat is in J/g*C, so we're good with that because g stands for grams and the aluminium is measured in grams. As there are 10 grams of aluminum, we have
10*(-33)*0.902=-298 ish as our final temperature

An exothermic reaction would release energy and would therefore lose heat itself, while an endothermic reaction would absorb energy and gain heat. Therefore, losing heat would be an exothermic reaction

Feel free to ask further questions!
4 0
3 years ago
Atoms A and X are fictional atoms. Suppose that the standard potential for the reduction of X^2+ is +0.51 V, and the standard po
IRINA_888 [86]

If you are given the standard potential for the reduction of X^2+ is +0.51 V, and the standard potential for the reduction of A^2+ is -0.33, just add the two. The standard potential for an electrochemical cell with the cell is 0.18V

5 0
3 years ago
Read 2 more answers
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
What is temperature?
BabaBlast [244]

the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch. So I would have to go with A.

8 0
3 years ago
Read 2 more answers
Show the difference in the reactivity of Na,Ca,Mg,Al,Fe,Cu with H2O
Vesnalui [34]

Answer:

Explanation:

Na react with H2O to form NAOH

2 Na+2H2O....................2NAOH + H2

Ca react with water and form calcium hydroxide

Ca + 2H2O........................Ca(OH)2

Mg react with water and form Magnesium hydroxide

Mg +2H2O .........................Mg(OH)2 however this coating of mg(oh)2 prevent it from further reaction

 Fe react with water and form ferric hydride

3Fe +H2O.......................2 FeH +FeO

copper do not react with water

7 0
3 years ago
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