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4 years ago

What is the theoretical yield for 3.14 g of magnesium, and .367 g of magnesium?

Chemistry

1 answer:

Dennis_Churaev [7]4 years ago

6 0

Answer:

Explanation:

i dont know

Magnesium is a chemical element with the symbol Mg and atomic number 12. It is a shiny gray solid which bears a close physical resemblance to the other five elements in the second column of the periodic .

Symbol: Mg

Atomic mass: 24.305 u

Electron configuration: [Ne] 3s2

Atomic number: 12

Melting point: 650 °C

Oxidation number: 2

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A student isolated 15.6 g of product from a chemical reaction. She calculated that the reactions should have produced 18.4 g of

Snezhnost [94]

Answer:

The percent yield of this reaction is 84.8 % (option A is correct)

Explanation:

Step 1: Data given

The student isolated 15.6 grams of the product = the actual yield

She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams

Step 2: Calculate the percent yield

Percent yield = (actual yield / theoretical yield ) * 100 %

Percent yield = (15.6 grams / 18.4 grams ) * 100 %

Percent yield = 84.8 %

The percent yield of this reaction is 84.8 % (option A is correct)

7 0

4 years ago

Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.

zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:

Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

Purple box

T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

$\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm$

Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

$\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC$

Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol$

3 0

3 years ago

A molecule of pentane is made of 5 carbon atoms and 12 hydrogen atoms. a collection of pentane molecules has 3.01 x 1024 carbon

Novay_Z [31]

3.01 Ă— 10^24 Ă— (12/5) hydrogen atoms
Looking at the formula for the molecule, the ratio of carbon to hydrogen atoms is 5:12, so if we divide the number of carbon atoms by 5 and then multiply by 12, we can find the number of hydrogen atoms. Let's look at the available options and see what makes sense.
3.01 Ă— 10^24 Ă— (12/5) hydrogen atoms
* This is exactly correct.
(3.01 Ă— 10^24 / 5) hydrogen atoms
* Nope. This will tell you how many pentane MOLECULES you have, but not the number of hydrogen atoms.
3.01 Ă— 10^24 Ă— (5/12) hydrogen atoms
* Close, but the ratio (5/12) will tell you the number of carbon atoms you have if you give it the number of hydrogen atoms. So this choice is wrong.
3.01 Ă— 10^24 Ă— 12 hydrogen atoms description
* This would tell you the number of hydrogen atoms you have if you know the number of pentane molecules you have. So this choice is also wrong.

5 0

4 years ago

Read 2 more answers

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

Purple as per the color of potassium permanganate

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$

7 0

2 years ago

sample of substance X has a mass of 326.0 g releases 4325.8 cal when it freezes at its freezing point. If substance X has a mola

ELEN [110]

Number of moles ( substance x ):

1 mole --------- 58.45 g/mol
? mole --------- 326.0 g

326.0 x 1 / 58.45 => 5.577 moles

heat of fusion:

hf = Cal / moles

hf = 4325.8 Cal / 5.577 moles

hf = 775.65 cal/mol

hope this helps!

5 0

3 years ago

Read 2 more answers