Answer:
<u>167.2 g</u>
Explanation:
Known
VC4H10 = 21.3
T = 0.00 C (convert to Kelvin: 273 K)
P = 1.00 atm
Unknown
m = ?g
1. <u>Write the balanced chemical equation</u>
1 C4H10 + 1O2 -----> 4 CO2 + 5 H2O
2. <u>Find the volume ratio of Carbon Dioxide to Butane </u>
1 C4H10 4 CO2 = 4 volumes CO2 / 1 volume C4H10
3.<u> Multiply by the known volume of n (butane)</u>
21.3 L C4H10 x 4 volumes CO2 / 1 volume C4H10 = 85.2 L C4H10
4. <u>Use ideal gas law</u>
PV = nRT solve for n ----> n = PV/RT
n= (1.00 atm) (85.2 L) / (0.0821 L atm/mol K) (273) = 3.80 mol CO2
5.<u> Find molar mass of CO2</u>
1 C x 12 + 2 O x 16 = 44.00
6. <u>Multiply the ideal gas law solution (3.80) by molar mass CO2 (44.00)</u>
3.80 mol CO2 x 44.00 g CO2
= 167.2 g CO2
Answer : The correct option is "record 10.00 ml in the notebook after using it."
Explanation :
If an instrument has "10 mL TD" written on it, that means the instrument is meant To Deliver (TD) 10 mL of volume.
Therefore the option that has 5.00 mL is not correct.
"+/- 0.01 mL" indicates the precision of the instrument. We know that no measurement is 100% accurate and there is always some uncertainty associated with any measurement.
Here, 0.01 indicates the uncertainty in the measurement.
When we have 0.01 mL , that means the instrument can record precisely to the hundredths place.
The more the significant figures, the greater is the accuracy of the measurement.
Therefore when we use an instrument that has "+/- 0.01 mL" written on it, we should record the reading as 10.00 mL
Based on the question, it is evident that the question is based on Boyle's Law which we will use to find the answer necessary.
Now, Boyle's Equation: P₁V₁ = P₂V₂
V₂ = P₁V₁ ÷ P₂
∴ V₂ = (12.5L × 42.0 cm · Hg) ÷ 75 cm · Hg
⇒ V₂ = 7 L
Thus, the <span>volume when the pressure has increased to 75.0 cm Hg is 7 Litres.</span>
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
Answer:
no
Explanation:
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