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3 years ago

the grams of carbohydrate in one apple if the apple has no fat and no protein and provides 72 kcal of energy

1 answer:

5 0

<span>Carbohydrates have 4 kcal per gram of carbohydrates. (Protein is also 4 kcal/gram, while fat is 9 kcal/gram). Take the total kcal and divide by 4 kcal/gram: 72 kcal/(4 kcal/gram) = 18 grams carbohydrates.</span>

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Why is oxygen crucial to living things?

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3 years ago

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3 years ago

Molarity of 2%W/V of NaOH is

TiliK225 [7]

Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
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3 years ago

How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq

Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

8 0

3 years ago

How many formula units make up 36.0 g of magnesium chloride (MgCl2)?

zalisa [80]

Answer: There are $2.29\times 10^{23}$ formula units

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{36.0g}{95g/mol}=0.38moles$

1 mole of $MgCl_2$ contains = $6.023\times 10^{23}$ formula units

Thus 0.38 moles of $MgCl_2$ contains = $\frac{6.023\times 10^{23}}{1}\times 0.38=2.29\times 10^{23}$ formula units

Thus there are $2.29\times 10^{23}$ formula units

6 0

3 years ago