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zimovet [89]
3 years ago
9

Look at the periodic table in Appendix 1. Name two elements that do not appear in order of their relative atomic masses.

Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0

iodine which should appear before tellurium and argon which should appear after potassium

There is a 3rd one: nickel should appear before cobalt

You might be interested in
Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip
erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0
3 years ago
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) <u>Mg is the limiting reactant</u>

<u>d) Mg is the limiting reactant</u>

<u>e) Nor Mg, neither I2 is the limiting reactant.</u>

<u>f) I2 is the limiting reactant</u>

<u>g) Nor Mg, neither I2 is the limiting reactant.</u>

<u>h) I2 is the limiting reactant</u>

<u>i) Mg is the limiting reactant</u>

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. <u>Nor Mg, neither I2 is the limiting reactant.</u>

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)

<u>I2 is the limiting reactant</u>, 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)

<u>Mg is the limiting reactant</u>, 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)

<u>Mg is the limiting reactant</u>, 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. <u>Nor Mg, neither I2 is the limiting reactant.</u>

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)

<u>I2 is the limiting reactant</u>, 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. <u>Nor Mg, neither I2 is the limiting reactant.</u>

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

<u>I2 is the limiting reactant</u>, 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

<u>Mg is the limiting reactant</u>, 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
The dots in these cylinders represent the shape and
Elden [556K]

Answer:

A I think

Explanation:

Because plasma have ions and B is neutralized with two ions

8 0
3 years ago
A pure substance where atoms of two or more elements are chemically bonded to one another
inn [45]
Compound is a pure substance where atoms of two or more elements are chemically bonded togather
8 0
3 years ago
In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0
3 years ago
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