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strojnjashka [21]
3 years ago
7

Answer the following question attached!!

Chemistry
1 answer:
irina [24]3 years ago
4 0

Answer:

I think It is Covalent. Not 100% sure but like 90% sure.

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What are the molarity and the normality of a solution made
Nikitich [7]

Explanation:

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,    Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

Hence, calculate the molarity of given solution as follows.

     Molarity of citric acid = \frac{\text{mass of citric acid}}{\text{molar mass of citric acid}} \times \frac{1}{\text{volume of solution(L)}}

                                 = \frac{25 g}{192.13 g/mol} \times \frac{1}{0.75 L}

                                 = 0.173 M

As citric acid is a triprotic acid so, upon dissociation it gives three hydrogen ions.

          Normality = Molarity × no. of hydrogen or hydroxide ions

                            = 0.173 × 3

                            = 0.519 N

Thus, we can conclude that molarity of given solution is 0.173 and its normality is 0.519 N.

3 0
3 years ago
At room temperature, a melting solid produces a whitish liquid. A laser beam shined through the liquid is unaffected and produce
navik [9.2K]
The answer is a colloid.
8 0
3 years ago
What are the nonmentals in group 16 of the periodic table?
Kitty [74]
I can give you four oxygen, selenium, sulfur, talarium hope this helps!
4 0
3 years ago
What if you love someone a lot but are to afraid to say it?
Serjik [45]

Answer:

Zodiac signs.

Explanation:

Cringe.

4 0
3 years ago
Read 2 more answers
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
3 years ago
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