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Tatiana [17]
3 years ago
14

Write balanced equations for each of the processes described below. (Use the lowest possible coefficients. Omit states-of-matter

.)
a. Chromium-51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture.
b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a β particle.
c. Phosphorus-32, which accumulates in the liver, decays by β-particle production.
Chemistry
1 answer:
Aleks [24]3 years ago
7 0

Answer:

d

Explanation:

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L is the midpoint of VC<br><br> LV = 9 find VC
oksian1 [2.3K]

Answer:

VC = 18

Explanation:

Since L is the midpoint and you have LV, you know that LC is also 9.

4 0
3 years ago
Which particle has a mass that is approximately equal to the mass of a proton
Leto [7]
The answer is neutron
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3 years ago
In what order are chemical elements arranged on the periodic table?
Gala2k [10]

Answer:

an increasng atomic number/ increasing proton count

Explanation:

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2 years ago
How does a bimetallic thermometer works
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It's made up of two different metals bonded together, which expand by different amounts as they heat up. As the temperature changes, the bimetallic strip curves more or less tightly (contracts or expands) and the pointer, attached to it, moves up or down the scale.

7 0
3 years ago
Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th
Vera_Pavlovna [14]

Answer:

[H^+]=0.000285

pH=3.55

Explanation:

In this, we can with the <u>ionization equation</u> for the hydrazoic acid (HN_3). So:

HN_3~~H^+~+~N_3^-

Now, due to the Ka constant value, we have to use the whole equilibrium because this <u>is not a strong acid</u>. So, we have to write the <u>Ka expression</u>:

Ka=\frac{[H^+][N_3^-]}{[HN_3]}

For each mol of H^+ produced we will have 1 mol of N_3^-. So, we can use <u>"X" for the unknown</u> values and replace in the Ka equation:

Ka=\frac{X*X}{[HN_3]}

Additionally, we have to keep in mind that HN_3 is a reagent, this means that we will be <u>consumed</u>. We dont know how much acid would be consumed but we can express a<u> subtraction from the initial value</u>, so:

Ka=\frac{X*X}{0.004-X}

Finally, we can put the ka value and <u>solve for "X"</u>:

2.2X10^-^5=\frac{X*X}{0.004-X}

2.2X10^-^5=\frac{X^2}{0.004-X}

X= 0.000285

So, we have a concentration of 0.000285 for H^+. With this in mind, we can calculate the <u>pH value</u>:

pH=-Log[H^+]=-Log[0.000285]=3.55

I hope it helps!

8 0
3 years ago
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