Speed = Distance divided by time
Speed = 150 km / 7200 s
Speed = 1 km / 48 s
Key concepts
Density
Mass
Volume
Concentration
Buoyancy
Water
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Answer:
Complete ionic:
.
Net ionic:
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water:
,
, and
. These three salts will exist as ions:
- Each
formula unit will exist as one
ion and one
ion. - Each
formula unit will exist as one
ion and two
ions (note the subscript in the formula
.) - Each
formula unit will exist as one
and two
ions.
On the other hand,
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite
,
, and
(three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each
formula unit will exist as only one
ion and one
ion. However, because the coefficient of
in the original equation is two,
alone should correspond to two
ions and two
ions.
Do not rewrite the salt
because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of
and two units of
. Doing so will give:
.
Simplify the coefficients:
.
Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.