Answer:
±0.005 g
Explanation:
The uncertainty depends on whether the measurement was obtained manually or digitally.
1. Manual
The minimum uncertainty is ±0.01 g.
It may be greater, depending on random or personal errors
2. Digital
Most measurements of mass are now made on digital scales.
A digital device must always round off the measurement it displays.
For example, if the display reads 20.00, the measurement must be between 20.005 and 19.995 (±0.005).
If the measured value were 20.006, the display would round up to 20.01.
If the measured value were 19.994, the display would round down to 19.99.
The uncertainty is ±0.005 g.
The scale shown below would display a mass of 20.00 g
Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO
Solving for X,
X = (56 g × 24.7 g) ÷ 88 g
X = 2.26 g ÷ 88 g
X = 0.0257 g of CO
Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
Answer:
The exposed metal rusts is an example of a chemical change because rust is an example of a chemical change in objects for example bicycles, scooters, etc.
Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
Answer:
The correct answer would be - 2.4KJ or, 2400J
Explanation:
Given:
heat capacity of liquid water - 4.18 J/g·°C
heat of vaporization - 40.7 kJ/mol
Mass of water = 1g
Moles of water = mass/molar mass
= 1g/18.016g
= 0.055 moles
Then,
Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)
= m *s*dt + moles * heat of vaporization
= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ
= 137.94J + 2.26KJ
=0.138KJ + 2.26KJ
=2.4KJ or, 2400J
Thus, the correct answer would be - 2.4KJ or, 2400J
