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2 years ago

URGENT

Chemistry

1 answer:

Vadim26 [7]2 years ago

7 0

Answer:

Explanation:

Discussion

When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?

N2(g) + 3H2(g) ⇌ 2NH3(g)

The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols

on the left side.

The right side has 2 mols of NH3 = 2 mols on the right.

Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.

Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.

2NH3 is only two mols.

The equilibrium shifts Right

Answer

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If an object floats in syrup, it will float in oil

Viefleur [7K]

No, it won't.
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3 years ago

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If aqueous solutions of sodium chloride and mercury(I) nitrate are mixed, which insoluble precipitate is formed?

NeX [460]

Answer:

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

Explanation:

The balanced equation for the reaction is given below:

2NaCl(aq) + Hg2(NO3)2(aq) —> 2NaNO3(aq) + Hg2Cl2(s)

Considering the states of each compound in the reaction, we can see that Hg2Cl2 is in solid form meaning it will precipitate out of the solution

In solution the following occurs:

NaCl —> Na+(aq) + Cl-(aq)

Hg2(NO3)2 —> Hg2^2+(aq) + 2NO3^-(aq)

Combining the two equation together, a balanced double displacement reaction occurs as shown below:

2Na+(aq) + 2Cl-(aq) + Hg2^2+(aq) + 2NO3^-(aq) —> 2Na+2NO3^-(aq) + Hg2^2+2Cl-(s)

From the above we can thus right the insoluble precipitate as

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

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3 years ago

Bess [88]

Answer:

4Al + 3O₂ → 2Al₂O₃

Explanation:

4Al + 3O₂ → 2Al₂O₃

Only this reaction above obeys the law of conservation of mass. The others flout the rule.

The law of conservation of mass states that "matter is neither created nor destroyed in a chemical reaction but are simply transformed from one form to another".

By this law, the number of atoms on both sides of expression must be the same;

Number of atoms

Elements Left hand side Right hand side

Al 4 4

O 6 6

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2 years ago

Hello! I am working on some homework and came across this question, which completely confuses me for various reasons. Experiment

Mariana [72]

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3 years ago

The work function for metallic magnesium is 3.68 eV. Calculate the velocity in km/s for electrons ejected from a metallic magnes

Lemur [1.5K]

Answer:

v = 301.62 km/s

Explanation:

Given that:

The work function of the magnesium = 3.68 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $3.68\times 1.6022\times 10^{-19}\ J=5.8961\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=315\ nm=315\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{315\times 10^{-9}}=5.8961\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$5.8961\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{315\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=6.31047619\times 10^{-19}-5.8961\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=0.41437619\times 10^{-19}$

$v^2=0.0909717212\times 10^{12}$

v = 3.0162 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 301.62 km/s

8 0

3 years ago