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pishuonlain [190]
3 years ago
15

The nucleus of an atom stays together only because the repulsive forces, called

Chemistry
1 answer:
Ahat [919]3 years ago
7 0

Answer:

Electrostatic

Explanation:

The forces that are overcome are the repulsive electrostatic forces between the protons (all charged positively).

You might be interested in
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
Given the two half-cell reactions for copper and iron below
liraira [26]

Answer:

No, i will not use a water pipe consisting of the two metals

Explanation:

Looking at the reduction potential of the both metals, it is clear that an electrochemical cell is set up with iron as the anode and copper as the cathode.

This will make the iron to quickly corrode and eventually destroy the water pipe. It is better to have a set up in which another metal that is higher than iron in the electrochemical series is combined with it.

6 0
3 years ago
How many grams of Mg are needed to produce 224 g of MgO in the complete reaction of Mg
zmey [24]

Answer:

134.4 g of Mg

Explanation:

reaction:

2Mg + O2 ➡️ 2MgO

1) find the mol of MgO

mol = mass / molar mass

mass = 224 g

molar mass = 24+16 = 40

mol = 224 / 40

= 5.6 moles

2 mol = 5.6 moles

2) find the mass of Mg

mass = mol × molar mass

mol = 5.6

molar mass = 24

mass = 5.6 × 24

= 134.4 g

7 0
3 years ago
BRAINIEST PLS HELPP
Akimi4 [234]

Answer:

0.179kg/m3 hope that helps you out

4 0
3 years ago
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