Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
Answer: yes they do.
Explanation:
I'm tryna get more points because my other account got blocked for the next 48 hours.
(Don't be mad at me please)
Atomic mass. Which is the number of protons and neutrons combined.
Answer:
4 atmospheric pressure is needed to get a volume of 25 litres
