Answer:
Density = 4.06 * 10^14
Explanation:
Givens
r = 1 * 10^-13 cm
m = 1.7 * 10^-24 grams
Formula
V = 4/3 * pi * r^3
D = m/V
Solution
Put the givens into the formula.
V = (4/3) 3.14 * (1 * 10^-13)^3 Find the value for r^3
V = (4/3) 3.14 * 1 * 10^-39 Find the volume. Indicate divide by 3
V = 1.256 10^-38/3
V = 4.1867 * 10^-39
Density = m / v
Density = 1.7*10^-24/4.1867*10^-39 Do the division
Density = .406 * 10^(-24 + 39)
Density = .406 * 10^15 grams/ cm^3
Answer: Density = 4.06 * 10^14
Answer:
For example, atoms in Groups 1 and 2 have 1 and 2 valence electrons, respectively. Atoms in Groups 13 and 18 have 3 and 8 valence electrons, respectively. Valence electrons are responsible for the reactivity of an element. They determine how "willing" the elements are to bond with each other to form new compounds.
Explanation: it might be 18 my guess:)
unit coversation
1.429 atm - 1086mmhg
9361 pa-9.36 KPa - 70.21 mmhg
725 mmhg -0.95 atm- 96.26 kpa
calculation
(a) 1 atm = 760 mmhg
1.429 atm = ?
1.429 x760/1 = 1086.34 mm hg
(B) 1 mmhg = 101.325 kpa
? =9361 KPa
9361 x1 /101.25 =70.21 mmhg
760 mm hg= 101.325 KPa
70.21 mm hg=?
70.21 x101.325/760 = 9.36 Kpa
(C ) 1 atm = 760 mmhg
? = 725
= 725 x1/ 760=0.95 atm
1 atm = 101.325 kpa
0.95 =?
0.95 x101.325/1 = 96.26 KPa
The temperature of the gas sample is 813 K.
<u>Explanation:</u>
We have to use the ideal gas equation to find the temperature of the gas sample.
The ideal gas equation is PV = nRT
Pressure, P = 429 mm Hg = 0.56 atm
Volume, V = 560 mL = 0.56 L
R = gas constant = 0.08205 L atm mol⁻¹K⁻¹
Mass = 0.211 g
Molar mass of carbon di oxide = 44.01 g / mol
Moles, n = 
= 0.0047 mol
Now, we have to plugin the above values in the above equation, we will get the temperature as,
T = 
= 813 K
So the temperature of the gas sample is 813 K.
