Solution : As they have uniform composition throughout they are considered as homogeneous mixture. Both samples are mixture of two metals (gold and palladium) thus are alloys.
Explanation:
Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.
As a result, formula of magnesium oxide is MgO and not
.
The student write the equation as
, it is not correct.
Therefore, given equation will be balanced as follows.

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass.
For this, we look at the atomic masses of the elements present in the compound.
Cu has an atomic mass of 63.546 amu
Fe has 55.845 amu
and S has 36.065 amu
Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu
So we have not established the mass of the compound in amus
63.546 + 55.845 + 72.13 = 191.521
That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845
So to get the percentage, or fraction of iron, we take 55.845 / 191.521
Which comes out to 29.15% by mass
Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
The water molecules will flow from b to a due to osmosis.
Osmosis is where water molecules will flow from a region of higher water potential to a region of lower water potential, through a selectively permeable membrane.
When the water molecule concentration is higher, it has a higher water potential top. Water potential is the tendency for them to flow to a lower region.
The net movement will stop until both sides of the solution has a same water potential.
Answer:
6.82 g H₂S
General Formulas and Concepts:
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.200 mol H₂S
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
6.818 g H₂S ≈ 6.82 g H₂S