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Licemer1 [7]
3 years ago
13

How many calories of energy will 20 grams of carbohydrates provide if 1 gram provides 4 calories?

Chemistry
2 answers:
dedylja [7]3 years ago
7 0

20 grams at 4 calories per gram.

20*4=80 Calories.

in the end, A is your answer

Bas_tet [7]3 years ago
6 0

Answer:

i already commented it but i want points anyways its 80 because 20x4 = 80. it said one gram equals 4 calories and there is 20 grams so basically 4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4

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What is the pH of a 0.083 solution of HCI
quester [9]

Answer:

The pH is 1.08,  being an acid pH

Explanation:

pH is a parameter used to measure the degree of acidity or alkalinity of a substance. The pH scale ranges from 0 to 14. Values ​​less than 7 indicate the acidity range and those greater than 7 indicate alkalinity or basicity. Value 7 is considered neutral. Mathematically, pH is the negative logarithm of the molar concentration of hydrogen or proton ions (H⁺) or hydronium ions (H₃O⁺).

pH= - log [H⁺]= - log [H₃O⁺]

A strong acid is an acid that completely dissociates into hydrogen ions and anions in solution. This implies that the initial concentration of acid is equal to the final concentration of H₃O⁺. This occurs with acid HCl.

So,  pH is calculated as:

pH= - log (0.083)

pH= 1.08

<u><em>The pH is 1.08,  being an acid pH</em></u>

8 0
3 years ago
Is bleach liquid starch? <br> Yes or No
Jobisdone [24]

Answer:

O it's not

Explanation:

Have a great day!

8 0
2 years ago
Question 3 of 10
PIT_PIT [208]

Answer:

the moluculer formula is the answer

Explanation:

4 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
Find the number of moles of water that can be formed if you have 138 mol of hydrogen gas and 64 mol of oxygen gas.
charle [14.2K]
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O

138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O

128 mol H₂O
6 0
3 years ago
Read 2 more answers
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