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3 years ago

An element has an atomic number of 78. the number of protons and electrons in a neutral atom of the element would be _____.

Chemistry

1 answer:

Maurinko [17]3 years ago

5 0

78 and 78. An atom of the element has 78 electrons and 78 protons.

The atomic number is 78, so the atom has 78 protons (+).

If the atom is neutral, it must contain 78 electrons (-).

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Which element have zero valency

choli [55]

The answer is Helium. hope its helpful (:

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3 years ago

During respiration, energy is retrieved from the high-energy bonds found in certain organic molecules. Which of the following, i

babunello [35]

Answer:

(A.) CO2, H2O

Explanation:

The chemical equation for respiration process is:

$C_{6}H_{12} O_{6}$ + $6O_{2}$ → $6CO_{2}$ + $6H_{2} O$ + ATP↑

Energy is released during the biochemical process in the organism's cells in form of ATP. Byproducts of the reaction are carbon dioxide and water molecules.

Let me know if you require any further assistance.

4 0

3 years ago

____________J/g. Degree C

abruzzese [7]

Answer:

0.417 J/g'C

Explanation:

8 0

3 years ago

What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11

lutik1710 [3]

The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m

4 0

3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

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