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klio [65]
2 years ago
15

The half-life of radium-226 is 1,600 years. It decays into radon-222. What fraction of the original amount of radium-226 in a sa

mple will still be radium
after 8,000 years?
A.1/8
B.1/4
c.1/32
D.1/16
Chemistry
2 answers:
Mnenie [13.5K]2 years ago
6 0

Answer:

Your answer will be C. 1/32

vfiekz [6]2 years ago
5 0

Answer:

C. 1/32

Explanation:

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Help me please omg I don’t know
Anna35 [415]

Answer:

5 1 2 4and 3 this is correct way

8 0
2 years ago
2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer soluti
lutik1710 [3]

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

5 0
3 years ago
Reactions rates for reactions occurring in solution can be increased by increasing the concentration of the solution. With gases
LUCKY_DIMON [66]
"Increase Pressure " is the right answer. if you need help , let me know
4 0
3 years ago
Which of these statements is true?
vichka [17]
<span>All molecules are made up of atoms. N2 O2 and H2 exist in molecule form and not in atomic form.
They exist freely in nature as molecules.
According to above explanation,
</span><span>B. Some elements found in nature exist as molecules, is the correct answer.</span>
3 0
3 years ago
Read 2 more answers
What is the concentration of H+ ions at a pH = 2?
erastova [34]

Answer:

0.01M = [H⁺]; 1x10⁻¹²M = [OH⁻]; Ratio is: 1x10¹⁰

Explanation:

pH is defined as -log [H⁺]

For a pH of 2 we can solve [H⁺] as follows:

pH = -log [H⁺]

2 = -log [H⁺]

10^-2 = [H⁺]

<h3>0.01M = [H⁺]</h3>

Using Keq of water:

Keq = 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 0.01M = [OH⁻]

<h3>1x10⁻¹²M = [OH⁻]</h3><h3 />

The ratio is:

[H⁺] / [OH⁻] = 0.01 / 1x10⁻¹² =

<h3>1x10¹⁰</h3>

7 0
2 years ago
Read 2 more answers
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