The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
Answer:
Magnesium Difluoride
Explanation:
Mg = magnesium
F = fluoride
F2 = difluoride (prefix "di" indicates 2)
Answer:
At -13
, the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-
![\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D)
where
and
are initial and final pressure respectively.
and
are initial and final volume respectively.
and
are initial and final temperature in kelvin scale respectively.
Here
,
,
,
and
Hence ![T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}](https://tex.z-dn.net/?f=T_%7B2%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7DT_%7B1%7D%7D%7BP_%7B1%7DV_%7B1%7D%7D)
![\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D%5Cfrac%7B%28210.0kPa%29%5Ctimes%20%281.30L%29%5Ctimes%20%28250K%29%7D%7B%28150.0kPa%29%5Ctimes%20%281.75L%29%7D)
![\Rightarrow T_{2}=260K](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D260K)
![\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D%28260-273%29%5E%7B0%7D%5Ctextrm%7BC%7D%3D-13%5E%7B0%7D%5Ctextrm%7BC%7D)
So at -13
, the gas would occupy 1.30L at 210.0 kPa.
Answer:
1 year- 1 mole
time in general- amount of matter
1 second- 1 atom/ particle
Answer:
71.92 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
T1 = 50°C = 50 + 273 = 323K
V1 = 105L
T2 = -25°C = -25 + 273 = 248K
P2 = 105.4 kPa
P1 = ?
V2 = 55.0 L
Using P1V1/T1 = P2V2/T2
P1 × 105/323 = 105.4 × 55/248
105P1/323 = 5797/248
0.325P1 = 23.375
P1 = 23.375 ÷ 0.325
P1 = 71.92 kPa