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3 years ago

The nuclides I-131 and I-133 are classifieds as

Chemistry

2 answers:

ruslelena [56]3 years ago

4 0

Answer:

C) isotopes of the same element

Explanation:

An isomer is molecule with the same atoms as other but with a different distribution (position, optic, spacial, geometrical, etc).

This applies to molecules and not to atoms. (options A and B can't be)

The isotopes are two atoms of the same element (equal proton number) but with different mass number. This is the case of I-131 and I-133. SO option C is correct.

Also, I-131/Xe-131 and I-133/Cs-133 are two pairs of isobars (two atoms of different elements with the same mass number).

sasho [114]3 years ago

3 0

Isotopes are two atoms with the same number of protons, but a different number of neutrons. Because they have the same number of proton they are atoms of the same element, but with different masses.
so, the answer is c.
i hope this will help you

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three students made multiple weighings of a copper cylinder each using a different balance. Describe the accuracy and precision

ch4aika [34]

I think that you have put up an incomplete question. However, i am answering the question based on my research and knowledge.

Lissa- accuracy and precision are both low
Lamont- accuracy and precision are definitely high
<span>Leigh Anne- accuracy is low but precision is definitely high.
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3 0

3 years ago

Identify whether each element is a halogen, a noble gas, or nonmetal only.

OLga [1]

Answer:

Astatine: Halogen

Nitrogen: Non-Metal

Krypton: Non-Metal, Noble Gas

Chlorine: Non-Metal

Sulfur: Non-metal

Explanation:

5 0

3 years ago

Match the career with the duties. 1. set goals and objectives for company corporate management 2. develop safety programs legal

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1. coporate manager

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7 0

3 years ago

Read 2 more answers

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago

How many milliliters of 0.0630 m edta are required to react with 50.0 ml of 0.0110 m cu2 ?

gizmo_the_mogwai [7]

Moles Cu+2 = M * V
= 0.05 L * 0.011 m
= 0.00055 moles

when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles

and when the Molarity of EDTa = 0.0630 M

∴ Volume of EDTA = moles / Molarity
= 0.00055 / 0.0630
= 0.0087 L = 8.7 L

4 0

3 years ago