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denpristay [2]
3 years ago
6

How many protons and neutrons does gamma radiation have? it’s for my project please

Chemistry
1 answer:
aliina [53]3 years ago
4 0

None

it is a form of electromagnetic radiation, meaning it does not have any protons or neutrons

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1 i i r 1 X 1 I A mm mm nu. Inchon-Mvim u m an
Setler [38]
The answer is that the pilot was tired of life and committed suicide with hundreds of passengers.
8 0
3 years ago
What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio
evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)

We can write the law of mass action for it:

Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change x due to the dissociation extent, we are able to rewrite it as shown below:

1.8x10^{-5}=\frac{x*x}{0.100-x}

Thus, via the quadratic equation or solve, we obtain the following solutions:

x_1=-0.00135M\\x_2=0.00133M

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0
3 years ago
Read 2 more answers
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
The one i selected is wrong!! pls help:(( only answer if ur right
Darina [25.2K]

Answer:

Lithium's mass number would decrease by 1

Explanation:

7 0
2 years ago
The reaction 2 ClO2(g) + F2(g) → 2 FClO2(g) is first-order in both ClO2 and F2. When the initial concentrations of ClO2 and F2 a
kupik [55]

Answer:

3. 75.0%

Explanation:

2 ClO2(g) + F2(g) → 2 FClO2(g)

First order with respect to ClO2 and F2.

This means the rate equation is given as;

Rate = k [ClO2][F2]

When the initial concentrations of ClO2 and F2 are equal?

Let's assume an initial value of 1 for both reactants, so rate equation is given as;

Rate = k * 1 * 1 = k

The rate after 25% of the F2 has reacted is what percent of the initial rate?

The concentration left of F2 is 75% ( 100% - 25%) = 0.75

Concentration of ClO2 remains 1.

So rate equation is given as;

Rate = k * 1 * 0.75 = 0.75 k

Comparing 0.75k and k.

This means our answer is;

3. 75.0%

6 0
3 years ago
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