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3 years ago

Helium and nickel are examples of ________. These

Chemistry

1 answer:

emmasim [6.3K]3 years ago

3 0

Answer:

Elements, in turn, are pure substances—such as nickel, hydrogen, and helium—that make up all kinds of matter.

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Molybdenum can form a wide series of halide compounds, including four different fluoride compounds. The percent by mass of molyb

azamat

The formula and names of the compounds are:

1. Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

1. Determination of the name and formula of the molybdenum fluoride having 63.0% of molybdenum.

Molybdenum (Mo) = 63.0%

Fluorine (F) = 100 – 63 = 37%

<h3>Formula =? </h3>

Mo = 63.0%

F = 37%

Divide by their molar mass

Mo = 63.0 / 96 = 0.656

F = 37 / 19 = 1.947

Divide by the smallest

Mo = 0.656 / 0.656 = 1

F = 1.947 / 0.656 = 3

Therefore,

Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Determination of the name and formula of the molybdenum fluoride having 56.0% of molybdenum.

Molybdenum (Mo) = 56.0%,

Fluorine (F) = 100 – 56 = 44%

<h3>Formula =? </h3>

Mo = 56%

F = 44%

Divide by their molar mass

Mo = 56 / 96 = 0.583

F = 44 / 19 = 2.316

Divide by the smallest

Mo = 0.583 / 0.583 = 1

F = 2.316 / 0.583 = 4

Therefore,

Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Determination of the name and formula of the molybdenum fluoride having 50.0% of molybdenum.

Molybdenum (Mo) = 50.0%,

Fluorine (F) = 100 – 50 = 50%

<h3>Formula =? </h3>

Mo = 50%

F = 50%

Divide by their molar mass

Mo = 50 / 96 = 0.520

F = 50 / 19 = 2.632

Divide by the smallest

Mo = 0.520 / 0.520 = 1

F = 2.632 / 0.520 = 5

Therefore,

Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Determination of the name and formula of the molybdenum fluoride having 46.0% of molybdenum.

Molybdenum (Mo) = 46.0%,

Fluorine (F) = 100 – 46 = 54%

<h3>Formula =? </h3>

Mo = 46%

F = 54%

Divide by their molar mass

Mo = 46 / 96 = 0.479

F = 54 / 19 = 2.842

Divide by the smallest

Mo = 0.479 / 0.479 = 1

F = 2.842 / 0.479 = 6

Therefore,

Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

Learn more: brainly.com/question/11185156

7 0

2 years ago

What is the initial temperature (in Celsius) of a 675 ml balloon if the

RoseWind [281]

Answer: 127 (3.s.f)

Explanation:

Boyle's law states that Volume x pressure = constant

Therefore let's first find the constant: 45 x 1900 = 85500

Then lets plug 675 ml and 85500 into the equation above:

675 x C = 85500

C = 127 (3 s.f.)

7 0

2 years ago

Your experiment requires 150 mL of 7.7 M NaOH. How many grams of NaOH will you need?

Elodia [21]

You have molarity and you have volume. Use the formula :
Molarity(M)= Moles(N)/Liter(L) to get the solution.
150 ml= .150 L
7.7 = N/.150
N=.1.155 moles of NaOH.
And since you know the moles, use the molar mass to figure out the grams.
<span> (40g/mol NaOH) x (1.155mol) =
46.2 g of NaOH.</span>

7 0

3 years ago

A compound that is composed of molybdenum (Mo) and oxygen (O) was produced in a lab by heating molybdenum over a Bunsen burner.

irakobra [83]

First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo

We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g

We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O

To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O

0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53

Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.

5 0

3 years ago

Read 2 more answers

A student made the Lewis dot diagram of a compound as shown. Mg is written with two dots shown on its top. Cl is written on the

blsea [12.9K]

Answer:

The dots were not properly located and arrows are not used in Lewis structures

Explanation:

If we intend to write a Lewis structure for a compound, that lewis structure must consist of only dots. These dots actually show the valence electrons on the outermost shell of the molecule.

We do not involve arrows when writing dot electron structures for compounds. The valence electrons of magnesium ought not to be written together because they are not a lone pair, rather they are two unpaired electrons. The use of an arrow suggests a coordinate covalent bond which is not the case here.

The correct lewis structure for MgCl2 is shown in the image attached to this answer.

7 0

3 years ago