Answer:
the change is evaporation
Explanation:
the water heats up at the surface of the water and evaporates
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Answer:
NaI > Na2SO4 > Co Br3
meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.
Explanation:
The freezing point depression is a colligative property.
That means that it depends on the number of solute particles dissolved.
The formula to calculate the freezing point depression of a solution of a non volatile solute is:
ΔTf = i * Kf * m
Where kf is a constant, m is the molality and i is the van't Hoff factor.
Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.
The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.
As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:
NH4 I → NH4(+) + I(-) => 2 ions
Co Br3 → Co(+) + 3 Br(-) => 4 ions
Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.
So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
Answer:
See the explanation
Explanation:
1) The Lewis structure for
has a central Carbon<em> </em>atom attached to Oxygen atoms.
In the
we will have a structure: O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.
2) These atoms are held together by <u>double bonds.</u>
<u></u>
Again in the structure of
: O=C=O we only have double bonds.
3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.
Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.
4. The carbon atom is <u>Sp</u> hybridized.
We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.
5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.
(See figures)
Figure 1: Carbon hybridization
Figure 2: Oxygen hybridization
Answer:
The mass of xenon in the compound is 2.950 grams
Explanation:
Step 1: Data given
Mass of XeF4 = 4.658 grams
Molar mass of XeF4 = 207.28 g/mol
Step 2: Calculate moles of XeF4
Moles XeF4 = mass XeF4 / molar mass XeF4
Moles XeF4 = 4.658 grams / 207.28 g/mol
Moles XeF4 = 0.02247 moles
Step 3: Calculate moles of xenon
XeF4 → Xe + 4F-
For 1 mol xenon tetrafluoride, we have 1 mol of xenon
For 0.02247 moles XeF4 we have 0.02247 moles Xe
Step 4: Calculate mass of xenon
Mass xenon = moles xenon * molar mass xenon
Mass xenon = 0.02247 moles * 131.29 g/mol
Mass xenon = 2.950 grams
The mass of xenon in the compound is 2.950 grams