<h3>
Answer:</h3>
23.459 g NaNO₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄
[Given] 24.14714 g Na₂SO₄
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂
Molar Mass of Na - 22.99 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol
Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We need 5 sig figs (instructed).</em>
23.4587 g NaNO₂ ≈ 23.459 g NaNO₂
Answer:
.125 M
Explanation:
.15 M/L * .125 L = .01875 moles
now dilute to 150 cc (by adding 25 cc)
.01875M / (150/1000) = .125M
Unstable isotopes occur when the strong force is unable to overcome the <span> <span>electrostatic force.</span></span><span>
There are no stable isotopes in the elements at the upper end of the periodic table, which clearly demonstrates the limit of the ability of the nuclear binding energy or the residual strong force, to overcome the electrostatic repulsion of all those protons in the nucleus.
</span>
Answer:
"Well, by definition,
molarity=
<u>moles of solute </u>
volume of solution"
Explanation:
THE LINES UNDER <u>moles of solute </u>MEANS ITS OVER volume of solution"
I THINK IM RIGHT PLZ NOT HATE ON ME IF IN WRONG
Answer:
The response can be defined as follows:
Explanation:
ATP is a hydrolysis energy-saving money of the cell. It is used to support the cell's endothermic processes.
![ATP\ +\ H_{2}O\ \rightleftharpoons \ ADP\ +\ P_{i}\\\\Q\ = \ \frac{[ADP][P_{i}]}{[ATP][H_{2}O]}\\\\ADP= 0.250 \ M\\\\P_i = 0.010 \ M\\\\ATP = 0.150 \ M\\\\H_2O = 55.55 \ M\\\\\Delta G \ =\ \Delta G^{o}\ +\ 2.303\ RT\ \log\ Q\\\\R = 8.314 \frac{J}{mol\ K}\\\\T = 298\ K\\\\](https://tex.z-dn.net/?f=ATP%5C%20%2B%5C%20H_%7B2%7DO%5C%20%5Crightleftharpoons%20%5C%20ADP%5C%20%2B%5C%20P_%7Bi%7D%5C%5C%5C%5CQ%5C%20%3D%20%5C%20%5Cfrac%7B%5BADP%5D%5BP_%7Bi%7D%5D%7D%7B%5BATP%5D%5BH_%7B2%7DO%5D%7D%5C%5C%5C%5CADP%3D%200.250%20%5C%20M%5C%5C%5C%5CP_i%20%3D%200.010%20%5C%20M%5C%5C%5C%5CATP%20%3D%200.150%20%5C%20M%5C%5C%5C%5CH_2O%20%3D%2055.55%20%5C%20M%5C%5C%5C%5C%5CDelta%20G%20%5C%20%3D%5C%20%5CDelta%20G%5E%7Bo%7D%5C%20%2B%5C%202.303%5C%20RT%5C%20%5Clog%5C%20Q%5C%5C%5C%5CR%20%3D%208.314%20%5Cfrac%7BJ%7D%7Bmol%5C%20K%7D%5C%5C%5C%5CT%20%3D%20298%5C%20%20K%5C%5C%5C%5C)