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3 years ago

Explain the trend in atomic radius as you progress up column 17

1 answer:

3 0

As you progress up any column on the periodic table the atomic radius gets smaller. Fluorine is a very small atom because it has very few electrons and Iodine is a very large atomic radius because it has many electrons.

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Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)

stepan [7]

Answer:

ionic equation : 3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

net ionic equation: 3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)

Ionic equations: Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

Net ionic equations: Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

6 0

3 years ago

A certain solid has a density of 8.0 g/cm3. if 4.0 g of this solid are poured into 4.00 ml of water, which drawing below most cl

Alexandra [31]

Answer is: the volume of water after the solid is added is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.

4 0

3 years ago

Precipitation _____. occurs equally all over the globe is the change in state from a gas to a liquid happens when ice changes in

LenKa [72]

<h3>Answer;</h3>

exceeds evaporation over land

Precipitation exceeds evaporation over land

<h3>Explanation;</h3>

In order to maintain earths water balance, evaporation exceeds precipitation over oceans but precipitation exceeds evaporation over land.
Water evaporates into the atmosphere from the ocean and to a much lesser extent from the continents. Winds transport this moisture-laden air, often great distances, until conditions cause the moisture to condense into clouds and to precipitate and fall.
Most precipitation originates by evaporation from the oceans. Over time, water evaporated from the oceans is replenished by inflow of freshwater from rivers and streams.

4 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$