<h3>
Answer:</h3>
0.0253 mol H₂O
<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>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 0.456 g H₂O (water)
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.025305 mol H₂O ≈ 0.0253 mol H₂O
Answer:
Explanation:
Given that:
the temperature
= 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:

where; B = -
C = -5800 
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have


Multiplying through with V² ; we have


V = 2250.06 cm³ mol⁻¹
Z = 
Z = 
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :












The compressibility is calculated as:


Z = 0.9386


V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At 
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z = 
Z = 
Z = 0.588
Answer:
3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

28%
Explanation:
mass of solute(KBr) = 3.73g
mass of solvent(H2O) = 131g
mass of solution = mass of solute + mass of solvent
= 3.73 + 131
= 134.73g

Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Thus, the concentration of NaOH is, 0.336 M