PLEASE HELP !!!

Use the ruler to determine the length of the line measured to the correct degree of precision. 13.2 cm 13.25 cm 130 cm 132 mm

lapo4ka [179]

Your answer would be 13.25cm

Hope that helps!!!

3 0

3 years ago

Read 2 more answers

Which statement describes ONLY the Quantum Mechanical model of the atom? A) Energy Is quantized B) Has discrete energy levels. H

gayaneshka [121]

C) Has sub levels within the energy levels. Lol forgot to put C

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) 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

Miguel’s family drove 357 miles on their weekend trip. Their car’s average gas mileage was 25.5 miles per gallon. How many gallo

ad-work [718]

Answer:

14 gallons

Explanation:

357 divided by 14 = 25.5 and if you check your answer 14 x 25.5 = 357 (i hope this is right)

6 0

2 years ago

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A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and 32 C.

lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁ = 928kpa

T₁ = 129°C

V₁ = 569L

P₂ = 319kpa

T₂ = 32°C

Unknown:

V₂ = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

take the units to the appropriate ones;

kpa to atm, °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁ = 928kpa gives 9.16atm

T₂ = 32°C gives 273 + 32 = 305K

T₁ = 129°C gives 129 + 273 = 402K

Input the values in the equation and solve for V₂;

$\frac{9.16 x 569}{402} = \frac{3.15 x V_{2} }{305}$

V₂ = 1255.4L

4 0

3 years ago