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3 years ago

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Chemistry

1 answer:

Lana71 [14]3 years ago

8 0

Answer:

Yes

Explanation:

Based on the graph, nuclear energy is one of the least contributors of CO2 emissions.

You would support this source of energy .

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A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden

Luda [366]

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$ reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

3 0

4 years ago

The reaction Ba(NO3)2(aq) + Na2SO4(aq) >> 2NaNO3(aq) + BaSO4(s) goes to completion because a

Helen [10]

Answer:

2. precipitate is formed

Explanation:

In this reaction barium nitrate and sodium sulfate solutions react to form a barium sulfate precipitate.Barium and sulfate ions will react to give barium sulfate precipitate where as the sodium and nitrate ions are spectator ions.

The net ionic equation after removing unchanged ions from each side of the full ionic equation will be;

Ba²⁺+ SO₄²⁻ → BaSO₄ (s)

3 0

3 years ago

Read 2 more answers

A sample of CaCO3(s) is introduced into a sealed container of volume 0.638 L and heated to 1000 K until equilibrium is reached.

JulsSmile [24]

Answer:

the mass of CaO present at equilibrium is, 0.01652g

Explanation:

$K_p = [CO_2]$ = 3.8×10⁻²

Now we have to calculate the moles of CO₂

Using ideal gas equation,

PV =nRT

P = pressure of gas = 3.8×10⁻²

T = temperature of gas = 1000 K

V = volume of gas = 0.638 L

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.k

$\frac{3.8 * 10^2 * 0.638}{0.0821 * 1000} \\= 2.95 * 10^-^4$

Now we have to calculate the mass of CaO

mass = 2.95 * 10 ⁻⁴ × 56

= 0.01652g

Therefore,

the mass of CaO present at equilibrium is, 0.01652g

7 0

3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago

What breaks and shatters easily?

Sergio [31]

Glass breaks and shatter easily.

4 0

4 years ago

Read 2 more answers