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lina2011 [118]
3 years ago
6

A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model

?
A.) Subtract the atomic number from the mass number.
B.) Add the atomic number and the mass number.
C.) Add the number of electrons to the number of protons.
D.) Subtract the number of electrons from the number of protons.
Chemistry
1 answer:
san4es73 [151]3 years ago
4 0

c is the correct answer

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What type of reaction is shown in the following chemical equation in NH3+HC1 NH4C1​
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Explanation:

combination reaction

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Calculate the molar solubility of silver chloride in 0.15 M sodium chloride.
Alex777 [14]

Answer:

S = 1.1 × 10⁻⁹ M

Explanation:

NaCl is a strong electrolyte that dissociates according to the following expression.

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.

We can find the molar solubility (S) of AgCl using an ICE chart.

      AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                         0            0.15

C                      +S              +S

E                        S            0.15+S

The solubility product (Ksp) is:

Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)

If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M

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3 years ago
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What is nitrogen state at 20 celcius
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4 years ago
Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.
Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

HCOOH\leftrightarrow H^++HCOO^-

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}

1.8*10^-^4=\frac{x^2}{c-x}

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

1.8*10^-^4=\frac{x^2}{0.12}

Let's solve this for x. Multiply both sides by 0.12

2.16*10^-^5=x^2

taking square root to both sides:

x=0.00465

Now, we have got the concentration of [H^+] .

[H^+] = 0.00465 M

We know that, pH=-log[H^+]

pH = -log(0.00465)

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Mixed Practice:
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Given

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4 0
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