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givi [52]
3 years ago
14

In 100 grams of sweet peas there are 14.5 carbohydrates,5.7 grams of sugars,5.1 grams of dietary fiber, 5.4 grams of protein and

0.4 grams of fat what is the percent composition of each?
Chemistry
1 answer:
Brums [2.3K]3 years ago
7 0
14.5 % carb
5.7% sugar
5.1% fiber
5.4% protein
0.4% fat
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Explain how you determine molar mass of Ca(No3)2
igor_vitrenko [27]

Answer:

164

1st step we will write desperate molar mass of each element

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Which units can be used to measure length or distance? Check all that apply.
motikmotik

Answer:

<u>M</u><u>eter,</u><u> </u><u>kilometer </u><u>&</u><u> </u><u>inch </u>- used to measure length or distance.

8 0
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15 POINTS!! does anyone know the answer to this? is it A?
Norma-Jean [14]

Answer:

b.2

Explanation:

7 0
2 years ago
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How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature? A. 51,3
expeople1 [14]
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:

E = m(ΔH)

where:
m = mass of substance
ΔH = latent heat of fusion or melting

According to data, the ΔH of mercury is approximately 11.6 kJ/kg.

E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J

The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
3 0
3 years ago
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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
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