<h3>
Answer:</h3>
251 mol Xe
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.51 × 10²⁶ atoms Xe
[Solve] moles Xe
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rule and round. We are given 3 sig figs.</em>
250.747 mol Xe ≈ 251 mol Xe
Answer:
they will form a compound
The mass of water formed is
<u><em>calculation</em></u>
Use the ideal gas equation to calculate the moles of NH3 and O2
that is Pv= n RT
where; P= pressure,
V= volume,
n = number of moles,
R=gas constant = 0.0821 l .atm/ mol.K
make n the formula of the subject by diving both side by RT
n = PV /RT
The moles of NH3
n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles
The moles of O2
=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles
write the reaction between NH3 and O2
4 NH3 + 5 O2 →4 No +6H2O
from equation above 0.766 moles of NH3 reacted to produce
0.766 x 6/4 =1.149 moles of H2O
0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20
since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles
mass of H2O = moles x molar mass
from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol
mass = 18 g/mol x 0.9408 moles= 16.93 grams
pH of the solution after 24. 00 ml of the hcl has been added is 12.87
millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL
.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40......-2.40.........+2.40.... +2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)
pOH = -log (OH^-). Then
pOH = -log (0.075)
pOH =1.1249
As we know,
pH + pOH = pKw = 14.00
pH=14-pOH
pH=14-1.1249
pH=12.87
<h3>
What is pH?</h3>
pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.
The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.
Learn more about pH:
brainly.com/question/491373
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