First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.
<span>134 ml
First, let's determine how many moles of oxygen we have.
Atomic weight oxygen = 15.999
Molar mass O2 = 2*15.999 = 31.998 g/mol
We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml
Since the density is 1.149 g/mol, we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2
Divide the number of grams by the molar mass to get the number of moles
0.17235 g / 31.998 g/mol = 0.005386274 mol
Now we can use the ideal gas law. The equation
PV = nRT
where
P = pressure (1.0 atm)
V = volume
n = number of moles (0.005386274 mol)
R = ideal gas constant (0.082057338 L*atm/(K*mol) )
T = Absolute temperature ( 30 + 273.15 = 303.15 K)
Now take the formula and solve for V, then substitute the known values and solve.
PV = nRT
V = nRT/P
V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm
V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm
V = 0.133987239 L*atm / 1.0 atm
V = 0.133987239 L
So the volume (rounded to 3 significant figures) will be 134 ml.</span>
Answer:
Explanation:
1)The charge of one electron is given by
1 e = - 1.6 * 10-19 C
Then – 1 C = 1 e / ( 1.6 * 10-19 )
= 6.25 * 1018 e
So one-coulomb charge has 6.25 * 1018 electrons
2)Let q1 and q2 be two charges separated by a distance r
Then q1 = - 40 µC = - 40 * 10-6 C
And q2 = 108 µC = 108 * 10-6 C
Answer:
germanium (Ge)
Explanation:
both belong to the same group