Answer:
B. The substance is in the liquid state.
Explanation:
The symbol (l) in a chemical equation represents that the substance is in liquid state which is one of the physical states of a chemical substance.
Answer:
luster means either sheen/glow or reflected light
Explanation:
luster is the appearance of mineral's surface in terms of light-reflective qualities
Answer:
P = 28.5 atm
Explanation:
PV = nRT => P = nRT/V
P = Pressure = ?
n = moles of sample = mass/molar mass = 26g/44g·mol⁻¹ = 0.619 mole CO₂
R = gas constant = 0.08206 L·atm/mol·K
T = temperature in Kelvin = (35 + 273)K = 308K
V = volume in Liters = 550 ml = 0.550 L
Pressure (P) = nRT/V
= (0.619 mole)(0.08206 L·atm/mol·K)(308K)/(0.550 L) = 28.5 atm
Answer:
6.79 g of phosphine can be produced
Explanation:
The reaction is this:
3H₂ + 2P → 2PH₃
We have the mass of the two reactants, so let's find out the limiting reactant, so we can work with the equation. Firstly, we convert the mass to moles (mass / molar mass)
6.2 g / 30.97 g/mol = 0.200 moles of P
4g / 2 g/mol = 2 moles of H₂
Ratio is 3:2.
3 moles of hydrogen react with 2 moles of P
Then, 2 moles of H₂ would react with (2 . 2)/ 3 = 1.3 moles of P.
We have only 0.2 moles of P, so clearly the phosphorous is the limiting reactant.
Ratio is 2:2. So 2 moles of P can produce 2 moles of phosphine. Therefore, 0.2 moles of P must produce the same amount of phosphine.
Let's convert the moles to mass ( mol . molar mass)
0.2 mol . 33.97 g/mol = 6.79 g
<h3>
Answer:</h3>
62.3 mol Ca
<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>
3.75 × 10²⁵ atoms Ca
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
62.2717 mol Ca ≈ 62.3 mol Ca
