Answer: option b) 1 dg
Explanation:
1) These are the different equivalences of those quantities:
a) 1 g = 100 cg ⇒ 1 cg = 0.01 g
b) 1 g = 10 dg ⇒ 1 dg = 0.1 g
c) 1 g = 1000 mg ⇒ 1 mg = 0.001 g
d) 1g = 10⁹ g ⇒ 1 ng = 10 ⁻⁹g
2) Now that you have all the masures in grams you can compare:
0.1g > 0.01g > 0.001g > 10 ⁻⁹g
3) So, the largest value is 0.1g which is 1 dg.
To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine.
Know we will convert each of these masses to moles by using the atomic masses:
58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F
Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:
Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4
So, the empirical formula of the compound XeOF₄
Answer:
Cohesive forces are greater than adhesive forces
Step-by-step explanation:
The attractive forces between water molecules and the wax on a freshly-waxed car (adhesive forces) are quite weak.
However, there are strong attractive forces (cohesive forces) between water molecules.
The water molecules are only weakly attracted to the wax, so the cohesive forces pull the water molecules together to form beads
.
The relation between density and mass and volume is
the dose required is 2.5 tsp
each tsp contain 5mL
So dose required in mL = 2.5 X 5 = 12.5 mL
the mass will be calculated using following formula
The mass of dose in grams will be 15.38 g
Answer:
1.33 x 10⁻³ mol/L
Explanation:
The solubility of a gas in a liquid is directly proportional to its partial pressure .This is know as Henry´s law and is given by the formula:
S(g) = Kh x P
where S = solubility of the gas
Kh = henry´s constant for the gas
P = partial pressure of the gas above solution
we are told to assume that the composition of the air in the tank is the same as on land, hence its partial pressure and we know nitrogen composition in air is 78 %, so the partial pressure of N₂ is:
P N₂ = 0.78 x 2.73 atm = 2.13 atm
Now we can calcualte the solubility as follows:
S N₂ = 6.26 x 10⁻⁴ mol / L·atm x 2.13 atm = 1.33 x 10⁻³ mol/ L
