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LuckyWell [14K]
3 years ago
9

What is the atomic number of an element that has seven protons and eight neutrons and seven electrons?

Chemistry
2 answers:
Dmitrij [34]3 years ago
7 0

Answer:

Nitrogen has an atomic number of 7 (Z=7) because it has 7 protons in its nucleus. Some nitrogen atoms have an atomic mass number of 15 (A=15). A is the number of neutrons plus protons in the nucleus. However, we already know that there are 7 protons

Darya [45]3 years ago
6 0

Answer:

Atomic number = Number of protons = Number of electrons

Atomic number = 7

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Glycerine is known as a negative catalyst.why?​
ioda

Answer:

The presence of 1-2% ethanol as catalyst, suppresses the oxidation of chloroform with oxygen to give a poisonous gas called phosgene. ... Here glycerol acts as negative catalyst. Criteria or characteristics of catalysts. i. The mass and chemical composition of catalyst should remain unchanged at the end of the reaction.

Explanation:

6 0
3 years ago
How many grams of nh3 can be produced from 2.86 mol of n2 and excess h2?
IRISSAK [1]
I think <span>3H2+N2==>2NH3</span>
7 0
3 years ago
Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
ohaa [14]

<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>) = 0.021 M.
Ka(HCN) =  4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻] / [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] = x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M - x).
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4 0
3 years ago
What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t
const2013 [10]

Answer:

C = 0.08M

Explanation:

molar mass of AlCl3

Al =27

Cl = 35.5

27+3(35.5) =133.5g/mol

n= mass/Molar mass

n =CV

CV = mass/molar mass

C x 500 x 10^-³ = 5/133.5

C x 500 x 10^-³ = 0.04

C = 0.04/500 x 10^-³

C = 0.08M

4 0
3 years ago
in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0
3 years ago
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