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3 years ago

I need help with this ASAP!!!

Chemistry

2 answers:

RideAnS [48]3 years ago

7 0

You need help with what?

lisabon 2012 [21]3 years ago

3 0

A is 1
B is 1
C is 1/1836

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A 4Kg rock is rolling 10m/s find it’s kinetic energy

Lubov Fominskaja [6]

KE=.5mv^2
M=mass
v=velocity
.5(4)(100)=200
That should be the answer.

5 0

3 years ago

The atomic radius of a silver atom is 1.44 x 10-10. Give the

professor190 [17]

Answer:

0.144 nm

Explanation:

Silver's electronic configuration is (Kr)(4d)10(5s)1, and it has an atomic radius of 0.144 nm.

5 0

3 years ago

A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added

Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = $\frac{1.085 X 100}{1.45}$ =74.8 %.

7 0

3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

2 years ago

2C3H7OH + 9O2 --> 6CO2 + 8H2O

UNO [17]

Answer 12.2g is the answer

4 0

3 years ago