Explanation:
To solve this question, we will use the Clayperon Equation:
P.V = n.R.T
where:
P = 101.28 kPa
1 atm = 101,325 Pa
x atm = 101,280 Pa
x = 1 atm
V = 37.058 L
n = we don't know
R = 0.082 atm.L/K.mol
T = -139.88 ºC = -139.88+273.15 = 133.27 K
1*37.058 = n*0.082*133.27
n = 0.29 moles
Answer: 0.29 moles
<h3>
Answer:</h3>
0.50 mol SiO₂
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
30 g SiO₂ (sand)
<u>Step 2: Identify Conversions</u>
Molar Mass of Si - 28.09 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of SiO₂ - 28.09 + 2(16.00) = 60.09 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig figs and round. We are given 2 sig figs.</em>
0.499251 mol SiO₂ ≈ 0.50 mol SiO₂
Answer:
6.2 calories
Explanation:
Data Given:
change in temperature = 20 °C
specific heat of gold = 0.031 calories/gram °C
mass of gold = 10.0 grams
Amount of Heat = ?
Solution:
Formula used
Q = Cs.m.ΔT
Where:
Q = amount of heat
Cs = specific heat of gold = 0.031 calories/gram °C
m = mass
ΔT = Change in temperature
Put values in above equation
Q = 0.031 calories/gram °C x 10.0 g x 20 °C
Q = 6.2 calories
So option A is correct = 6.2 calories
Answer:
C.
Explanation:
Iron is most likely to form a precipitation reaction.