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2 years ago

If an element has 11 protons and 11 neutrons in its nucleus it must be ?

Chemistry

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

the answer is 11

Explanation:

Ok, let us understand what is an atomic number, an atomic number is simply the proton number, the atomic number is always seen at the bottom left corner in a element of a periodic table, it is the number of proton in a atom of a certain element.

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Write the net ionic equation for the acid-base hydrolysis equilibrium that is established when ammonium perchlorate is dissolved

Whitepunk [10]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which do not get involved in the chemical equation. It is also defined as the ions which are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium perchlorate and water is given as:

$NH_4ClO_4(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+HClO_4(aq.)$

Ionic form of the above equation follows:

$NH_4^+(aq.)+ClO_4^-(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+H^+(aq.)+ClO_4^-(aq.)$

Ammonium hydroxide will not dissociate into its ions because it is a weak base.

As, chlorate ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$NH_4^+(aq.)+H_2O(l)\rightarrow NH_3^+(aq.)+H_3O^+(aq.)$

Hence, the net ionic equation is given above.

5 0

3 years ago

Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g

ExtremeBDS [4]

Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

$gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl$

Have a nice day!

4 0

2 years ago

Which layer in the diagram was the original layer?

tatiyna

I’m almost sure it’s D but i could be wrong

4 0

1 year ago

Read 2 more answers

Can calcium conduct electricity

postnew [5]

Yes because it is a perfect conductor and all metals conduct heat.

8 0

2 years ago

Given the following information regarding the masses and relative abundances for the isotopes of an element, determine its atomi

Ilia_Sergeevich [38]

Answer:

Average atomic mass = 58.51047 amu

The symbol is $Ni$

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})+(\frac {\%\ of\ the\ fourth\ isotope}{100}\times {Mass\ of\ the\ fourth\ isotope})+(\frac {\%\ of\ the\ fifth\ isotope}{100}\times {Mass\ of\ the\ fifth\ isotope})$

Given that:

For first isotope:

% = 68.274 %

Mass = 57.9353 amu

For second isotope:

% = 26.095 %

Mass = 58.9332 amu

For third isotope:

% = 1.134 %

Mass = 61.9283 amu

For fourth isotope:

% = 3.593 %

Mass = 63.9280 amu

For fifth isotope:

% = 0.904 %

Mass = 63.9280 amu

Thus,

$Average\ atomic\ mass=\frac{68.274}{100}\times {57.9353}+\frac{26.095}{100}\times {58.9332}+\frac{1.134}{100}\times {61.9283}+\frac{3.593}{100}\times {63.9280}+\frac{0.904}{100}\times {63.9280}$

$Average\ atomic\ mass=39.55474 +15.37861854+0.702266922+2.29693304+0.57790912$

Average atomic mass = 58.51047 amu

The symbol is $Ni$

5 0

3 years ago