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3 years ago

Which statement describes the dependent variable?

1 answer:

6 0

The statement which describes the dependent variable would be the measured variable. We usually use the independent variable and manipulate it and change its levels to see how the levels of the measurable (dependent) variable change.

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What element does HgNo

melisa1 [442]

Answer:

Mercury is a chemical element with symbol Hg and atomic number 80. Classified as a transition metal, Mercury is a liquid at room temperature.

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Explanation:

3 0

2 years ago

Read 2 more answers

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

Name the following alkyne. Please help me <3

Alexandra [31]

Answer:

D. 7-methyl-3-octyne

Explanation:

1) Identify the parent chain and name it like an alkane.

• The longest chain of carbons, which consists of the functional group (which is the alkyne group in this case: C≡C).

• There are 8 carbons in the longest chain, so it is called an octane.

2) Now, identify the location of the functional group.

• The location number of the functional group should be as low as possible. Thus although we could count from the right, we start counting from the left since the functional group is closer to the left. From the left, the functional group would be at carbon 8 while from the left, it is on carbon 3.

• Replace 'ane' with 'yne' in octane for the alkyne group.

• This would give us 3-octyne.

3) Lastly, add in the name of the branch.

• Here we have one branch, -CH₃. This is read as methyl.

• Identify the location number of the branch by counting the number of carbons in the same direction as when we counted the location number of the functional group. The methyl branch has a location number of 7.

• Adding the name of the branch before the parent chain, we would arrive at 7-methyl-3-octyne as the IUPAC name of the alkyne.

Further Explanation:

A) This option is incorrect as there are only 8 carbons in the parent chain. Although there are 9 carbons in total, the 9th carbon is taken care of in '7-methyl'.

B) Location number of the functional group should be as low as possible, so start counting the number of carbons from the left!

C) Since the functional group is an alkyne, the word 'octane' should be 'octyne' instead.

8 0

3 years ago

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

ZanzabumX [31]

The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg.

7 0

3 years ago

How many “p” orbital in principal energy level below?

marshall27 [118]

Answer:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. Therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

Explanation:

5 0

3 years ago