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3 years ago

____ can be found in many forms including in the chemicals found in food

Chemistry

1 answer:

Rzqust [24]3 years ago

4 0

A. Chemical energy can be in many forms including in the chemicals found in food

You might be interested in

What mass of sodium chloride contains 4.59 x 10 24 formula units?

shtirl [24]

The term formula units means molecules.

Then, what you are looking for is the mass in 4.59*10^24 molecules.

The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.

1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol

2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

3) mass of NaCl = molar mass * number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl

Answer: 445.31 g of NaCl.

7 0

3 years ago

Given 1.00 grams of each of the following radioisotopes, Ca-37, U-238, P-32 and Rn-222,which would have the least remaining orig

erica [24]

1) Ca-37, with a half-life of 181.1(10) ms.

5 0

3 years ago

Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr

kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm 5.1 atm 0 Initial

-2x -x +2x Reacts (stoichiometry is 2:1:2)

4.9-2x 5.1-x 2x Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0

3 years ago

How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -><br> 4A1 + 302

Evgesh-ka [11]

<h3>Answer:</h3>

$\displaystyle 40 \ mol \ Al$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 40 \ mol \ Al$

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0

3 years ago

You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

Jobisdone [24]

You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)

4 0

3 years ago