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3 years ago

Which best describes the changes occurring to the earth's surface over time?

Chemistry

1 answer:

uysha [10]3 years ago

7 0

Answer:

I would say the answer is 2

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Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

Dmitrij [34]

Answer :

(a) The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

The spectator ions are, $NH_4^{+}\text{ and }SO_4^{2-}$

(b) The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

(c) The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)$

The ionic equation in separated aqueous solution will be,

$2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)$

In this equation, $NH_4^{+}\text{ and }SO_4^{2-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

(b) The given balanced ionic equation will be,

$Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

(c) The given balanced ionic equation will be,

$Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)$

The ionic equation in separated aqueous solution will be,

$Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

3 0

3 years ago

Which of the following will have the highest boiling point?

DerKrebs [107]

Answer:

0.1M NH3

Explanation:

The boiling point of aqueous solutions depend on the nature of intermolecular interactions present. KBr will yield an ionic solution but NH3 will yield a molecular solution having hydrogen bonds. The degree of hydrogen bonding in the aqueous solution will further increase with the concentration of the solution.

Remember that experimental data shows that hydrogen bonds are strong bonds that lead to a significant increase in the boiling point of solutions. Hence 0.1M NH3 solution will have a higher boiling point due to intermolecular hydrogen bonding in the solution.

8 0

3 years ago

an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with

ryzh [129]

Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

7 0

3 years ago

If the pressure of a gas sample is quadrupled and the absolute temperature is doubled, by what factor does the volume of the sam

Natasha_Volkova [10]

Answer:

D) 1/2

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given,

P₂ = 4P₁

T₂ = 2T₁

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{4\times P_1}\times {V_2}}{2\times T_1}$

$V_2=\frac{1}{2}\times V_1$

The volume change by half of the original.

7 0

3 years ago

Read 2 more answers

Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which

prohojiy [21]

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

Step 1: Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

Step 2: Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

Step 3: Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

8 0

3 years ago