Does anyone know where i can find a practice cumulative chem test online?

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

The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins

saul85 [17]

The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

So, we know:
t = 10 min
 $t_{1/2}$ = 2.5 min

We need:
n = ?
x = ?

We could first use the second equation to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{10 min}{2.5 min}$
⇒ $n=4$ 

Now we can use the first equation to calculate the remained fraction of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^4$
⇒ $x= \frac{1}{16}$

3 0

3 years ago

Hello sunrays~ help me with these questions

andre [41]

Explanation:

During the formation of BeCl2, the beryllium atom bonds with two chlorine atoms via single covalent bonds. The number of electron pairs around the central atom will be two. No lone pair is found in the molecule. If we analyse this information then we can conclude that BeCl2 has sp hybridization.

the cathode

4 0

2 years ago

Read 2 more answers

Describe the concept of using radioactive elements to establish an age for a fossil. Be sure to include how the parent and daugh

Nadusha1986 [10]

Answer:

The nuclear decay of radioactive elements is a process that is a useful tool for determining the absolute age of fossils and rocks. It is used as a clock, in which daughter elements or isotopes converted from parent isotopes by decaying at a particular time.

Radioactive decay rates are constant and do not change over time. It is measured in half-life. A half-life is a time it takes half of a parent isotope to decay and converted into a stable daughter isotope. How many parent isotopes and daughter isotopes present in the fossil or their abundance can help in determining the age of fossil or rock.

4 0

3 years ago

What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?

podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then 1.2 x 10²⁶ molecules will give:

$\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times 10^{23}}$

= 17.934 x 10³ g of water

= 17.934 kg of water

5 0

3 years ago