Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
A
Explanation:
The mallet transfers kinetic energy, then that energy is transferred to the orange ball, then the purple ball, thus making the purple ball move
Answer:
6.68 x 10^-4
Explanation:
131g ÷ 261.337g/mol = 0.5012685 moles
0.5012685 moles ÷ 750.0 liters =
0.5012685÷ 750.0=0.000668358
6.68 x 10^-4
<h3>
Answer:</h3>
10.6 mol NO
<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>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O
[Given] 13.2 mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 5 mol O₂ → 4 mol NO
<u>Step 3: Stoich</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>
10.56 mol NO ≈ 10.6 mol NO
