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Alisiya [41]
3 years ago
15

What are the correct two steps, in the correct order, that would allow you to solve for x in the following equation?

Chemistry
1 answer:
Pani-rosa [81]3 years ago
5 0

Answer:

4th one is correct...............

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Give the name and formula of the compound formed from the following elements:<br> (b) ₃₀L and ₁₆M
madam [21]

When sulfur reacts with zinc metal is forms a compound known as zinc sulfide.

<h3>Name of the elements</h3>

The names of the elements that make up the compound will be determined from the atomic mass of the elements.

<h3>Which element has atomic number 30</h3>

The element that atomic number of 30 is zinc and it has oxidation number of +2.

<h3>Which element has atomic number 16</h3>

The element that atomic number of 16 is sulfur and it has oxidation number of -2.

<h3>Chemical reaction between sulfur and zinc metal</h3>

Zn²⁺ + S²⁻  --------> ZnS

Thus, when sulfur reacts with zinc metal is forms a compound known as zinc sulfide.

Learn more about zinc sulfide here: brainly.com/question/20806552

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1 year ago
True or False: As the ice warms up, the molecules get more energy and move around more from place to place to form liquid water.
Arada [10]
The answer is TRUE.

Hope it helps.
7 0
3 years ago
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Sladkaya [172]

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3 0
3 years ago
Of the elements fe li te u and he which are considered group b elements
VLD [36.1K]
A group B element is Fe 
4 0
3 years ago
a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter
ladessa [460]

Answer:

The concentration of chloride ion is 2.82\times10^{-3}\;mol/L

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the Cl^- content 100 ppm suggests the presence of 100 mg of Cl^- in 1 L of solution.

The molar mass of Cl^- is equal to the molar mass of Cl atom as the mass of the excess electron in Cl^- is negligible as compared to the mass of Cl atom.

So, the molar mass of Cl^- is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of Cl^- present in 100 mg (0.100 g) of Cl^- is calculated as shown below:

N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol

So, there is 2.82\times10^{-3}\;mol of Cl^- present in 1 L of solution.

5 0
4 years ago
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