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3 years ago

What is the relationship between the number of protons and the number of electrons in an atom?

Chemistry

1 answer:

aleksley [76]3 years ago

6 0

In a neutral atom they are both equal, and their even quantities makes the atom neutral...

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What information is presented inside each square of the periodic table?

Verdich [7]

The symbol, the atomic mass, the number of protons and electrons

8 0

3 years ago

Determination of your estimate determines the BLANK and BLANK of a piece of equipment.

Zina [86]

Answer:

scalpel,mask

Explanation:

4 0

2 years ago

In the second step of this reaction, isocyanic acid reacts to form melamine and carbon dioxide: HNCO(l)→C3N3(NH2)3(l)+CO2(g) Bal

kykrilka [37]

Answer:

The coefficients are 6, 1, 3

Explanation:

HNCO →C3N3(NH2)3 + CO2

From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:

6HNCO → C3N3(NH2)3 + CO2

Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:

6HNCO → C3N3(NH2)3 + 3CO2

Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.

The coefficients are 6, 1, 3

8 0

3 years ago

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

3 years ago

An aqueous solution is found to be 3.64% potassium nitrate by mass, how many grams of solution should I evaporate the water from

Sidana [21]

Answer:

513.74 g of solution

Explanation:

% Mass grams are defined as the grams that are dissolved in salt (in this case, it would be potassium nitrate) dissolved every 100 g of the solution. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.

The relationship is:

3.64 g of potassium nitrate _____ 100 g solution

18.7 g of potassium nitrate _____ X = 513.74 g of solution

Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution

So, I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it.

8 0

3 years ago