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balandron [24]
3 years ago
8

Ms. Salis explains to her class that sodium (Na) and chlorine (Cl) atoms bond to make table salt

Chemistry
1 answer:
larisa [96]3 years ago
6 0
The answer is D a compound :)
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Is glass breaking physical or a chemical
serg [7]

Answer:

<h2>physical </h2>

Explanation:

I hope this helps

3 0
3 years ago
A certain chemical reaction releases 31.0 kj/g of heat for each gram of reactant consumed. How can you calculate what mass of re
Aleksandr-060686 [28]

Answer:

See explanation.

Explanation:

Hello!

In this case, since we know the heat of reaction per gram of reactant and we should know the total energy of reaction, but it is not there, we are going to assume it is 1200 J as usual in these problems, so you can change it to whatever your given heat is.

In such a way, we set up the math as shown below:

m= 1200J*\frac{1kJ}{1000J}*\frac{1g}{31.0kJ}

Which results:

m=0.0387g

Best regards!

7 0
3 years ago
If 13 grams of copper sulfate is reacted with zinc how much of each product is produced?
Sladkaya [172]

Answer:

No. Of Moles of zinc = m/Ar

= 13/ 65.38 = 0.198 moles

From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1

So only 0.198 moles of CuSO4 reacts, it is in excess

Mass = no of Moles X Mr

Mass = 0.198 X 159.5 = 31.59 grams

Volume = mass m denisty

Volume j 31.59 / 3.6 = 8.78 ml

Explanation:

i think this wrong

7 0
3 years ago
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
Which part of an atom is mostly empty space?
Brilliant_brown [7]
I’m pretty sure it’s B

Hope that helps

7 0
3 years ago
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