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Sergeeva-Olga [200]
3 years ago
15

A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is s

aying 91g I keep getting 175g. Can someone please explain?
Chemistry
1 answer:
Effectus [21]3 years ago
6 0

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

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I would say the answer is A.

5 0
3 years ago
A certain hydrocarbon has a molecular formula of C9H16. Which of the following is not a structural possibility for this hydrocar
snow_tiger [21]

Answer: Option (d) is the correct answer.

Explanation:

It is given that molecular formula is C_{9}H_{16}. Now, we will calculate the degree of unsaturation as follows.

Degree of unsaturation = C_{n} - \frac{\text{monovalent}}{2} + \frac{\text{trivalent}}{2} + 1

                                  = 9 - \frac{16}{2} + 1

                                  = 9 - 8 + 1

                                  = 2

As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.

Yes, this compound could be an alkyne as for alkyne D.B.E = 2.

But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.

Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.

6 0
4 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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stiv31 [10]

Answer:

2KClO3 》》2KCl +3O2

C+ O2》》CO2

number of C moles

Required O2 moles (According to the mole ratio )

Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles

Now you can find the mass of KClO3

I mentioned the useful steps which can guide you to get the answer.

Explanation:

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ipn [44]

Answer:

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Explanation:

3 0
3 years ago
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