Ask question

Ask question

All categories

2 years ago

4

A Snickers bar contains 27.0 g of sugars, 4.0 g of protein and 12.0 g of fat. How many food calories (Cal) are contained in the

bar? Assume that metabolizing 1 g of protein releases the same amount of energy as 1 g of carbohydrates.

1 answer:

Lapatulllka [165]2 years ago

5 0

Answer:

232 kcal

Explanation:

Data provided in the question:

Contents of the snickers bar

27.0 g of sugars

4.0 g of protein

12.0 g of fat

Now,

we know

1 gram of sugars = 4 kcal

1 gram of proteins = 1 gram of carb. = 4 kcal

1 g of fat = 9 kcal

Therefore,

Total calories of snickers bar = ( 27 × 4 ) + ( 4 × 4 ) + ( 12 × 9 )

= 108 + 16 + 108

= 232 kcal

You might be interested in

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

2 years ago

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

8 0

2 years ago

What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c

brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0

2 years ago

Read 2 more answers

Which of the following solution is more dilute and explain why?a)1M b)2M c)0.1M or d)0.009M

Taya2010 [7]

I can’t access the pictures. Sorry!

6 0

2 years ago

Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the

padilas [110]

Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

6 0

2 years ago