Answer:
1 mole of substance to be 6.022 × 1023
Explanation:
thats all i could come up with
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol
moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2
Formula of substance:
C₂H₄
Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
The density of the rock is 3.314g/mL
CALCULATE DENSITY:
- According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.
- This means that the volume of the rock can be calculated as follows:
Volume of rock = 62mL - 55mL
Volume of rock = 7mL
Density can be calculated using the formula as follows:
Density = mass ÷ volume
Density = 23.2 ÷ 7
Density = 3.314g/mL
Therefore, the density of the rock is 3.314g/mL
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Answer: 3.59
Explanation:
(2.06)(1.743)(1.00)
2.06 × 1.743 × 1.00
= 3.59058
Two of the multiplied digits are represented in 3 significant figures. Therefore, for correct representation, the result of the product should be written to three significant figures.
3.59058 to 3 significant figures:
First three digits = 3.59
Fourth digit '0' is less than 5, and thus rounded to 0 with other succeeding digits
Therefore, (2.06)(1.743)(1.00) to 3 significant figures equals :
3.59
Answer:
See the answers below
Explanation:
1) 100. mL of solution containing 19.5 g of NaCl (3.3M)
2) 100. mL of 3.00 M NaCl solution (3 M)
3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)
4) Number 1 and 5 have the same concentration (1.5M)
MW of NaCl = 23 + 36 = 59 g
For number 3
59 g ------------------- 1 mol
19,5 g ----------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For number 4,
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For number 5
Molarity (M) = 0.450/0.3 = 1.5 M
