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Sonbull [250]
3 years ago
11

Uncooked lean ground beef can contain up to 12% fat by mass. How many grams of fat would be contained in 0.97 lbs of ground beef

?

Chemistry
1 answer:
klasskru [66]3 years ago
8 0

52.8  grams of fat would be contained in 0.97 lbs of ground beef

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When Acid Rain falls on limestone, the rock is eroded and carbon dioxide is produced.
Sav [38]

<em>The formula of calcium carbonate is CaCO3 </em>

<em>The formula of nitric acid is HNO3. </em>

<em>When put together:</em>

<em>CaCO3 + HNO3 </em><em>= </em><em>Ca(NO3)2 + CO2 + H2O </em>

<em> The balanced equation:</em>

<em> CaCO3 + 2HNO3</em><em> = </em><em>Ca(NO3)2 + CO2 + H2O </em>

<em> </em>

<em></em>

6 0
3 years ago
After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage
Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide (C_6H_5Br)= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and  C_6H_5Br= 1:1

2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg

9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br

The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

4.83~g~\frac{1~mol}{136.14~g}=0.35~mol

The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:

\frac{0.35~mol}{1}=0.35

\frac{0.0402~mol}{2}=0.0201

Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:

Molar mass of triphenyl carbinol= 260.33 g/mol

0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol

Finally, we have to <u>divide</u> the obtanied solid by the calculated one:

Percentage=\frac{5}{5.23}*100=95.6\%

5 0
3 years ago
100 POINTS!!!!<br><br> Please I need this page done lol
Alenkinab [10]

Answer:

okay.. Questions?????????

5 0
3 years ago
Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.
OlgaM077 [116]

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

3 0
3 years ago
Read 2 more answers
Calculate the volume which 1.00 mole of a gas occupies at 1 atm and 298K?
elena-14-01-66 [18.8K]

Answer:

25.45 Liters

Explanation:

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (1mole)(0.08206Latm/molK)(298K)/(1atm) = 25.45 Liters

7 0
3 years ago
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