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noname [10]
3 years ago
6

Calculate the number of calories in a food product that has 14g of fat, 20g of carbohydrates and 4g of protein.

Chemistry
1 answer:
KonstantinChe [14]3 years ago
5 0
20..................................
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What is the molality of a solution of 12.9 g of fructose (C6H12O6) in 31.0 g of water?
jasenka [17]

Answer:

2.31 m C₆H₁₂O₆

Explanation:

Hope this helps

8 0
3 years ago
A student measures the volume of a solution to be 0.01370 l. How many significant digits are in this measurement?
Mkey [24]

A student measures the volume of a solution to be 0.01370 have 5 significant digits in this measurement.

<h3>What are significant digits?</h3>

The significant digits are the minimum number from zero to nine for reporting any measurement where the digits are uncertain.

The significant digits starting from zero are not significant digits, decimal is not a significant digit, and ending zero after the decimal are significant digits.

Therefore, the student measures the volume of a solution to be 0.01370 5 significant digits are in this measurement.

Learn more about significant digits, here:

brainly.com/question/1658998

#SPJ4

7 0
2 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
Which is an example of a compound?
makkiz [27]

Answer:

Octane - Formula: C8H^18 = Carbon^8 + Hydrogen^18

Explanation: Octane is a compound because there are 8 atoms of carbon and 18 atoms of hydrogen in one molecule of C8H18. There are also 8 moles of carbon and 18 moles of hydrogen.

5 0
3 years ago
The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)
natita [175]

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

8 0
4 years ago
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