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3 years ago

Need Help Please

Chemistry

1 answer:

sergejj [24]3 years ago

3 0

Answer:

A positive ion has more protons than electrons and a negative ion has more electron ions

So the answer is

A positive ion has less electrons and a negative ion has more electrons

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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?

I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄ + 2H₂O₂ → N₂ + 4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

N₂H₄ : H₂O₂

1 : 2

1.07 : 2×1.07 = 2.14 mol

6 0

3 years ago

Sulfurous acid reacts with barium hydroxide.

Ivenika [448]

Answer:

Ba(OH)2 + H2SO3 = BaSO3 + H2O -

Explanation:

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3 years ago

The first step of electrophilic aromatic substitution involves the formation of the arenium ion intermediate.

Anastaziya [24]

Answer:

True

Explanation:

Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.

Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.

4 0

3 years ago

For the reversible, one-step reaction A+B<---->C+D, the forward rate constant is 52.4 /mol*h and the rate constant for the

Fantom [35]

Answer:

—-—1———————-s

7 0

3 years ago

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.600 M. N2(g)+O2(g)↽−−⇀2NO(g) If

cestrela7 [59]

Answer:

0.84M

Explanation:

Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

$K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9$

Now, the new equilibrium condition, taking into account the change x, becomes:

$9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}$

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

$\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}$

Thus, we arrange the equation as:

$\frac{1}{9} (0.9-2x)^2=(0.2+x)^2\\0.09-0.4x+4x^2=0.04+0.4x+x^2\\3x^2-0.8x+0.05=0\\x_1=0.06$

Finally, the new concentration is:

$[NO]_{eq}=0.9-0.06=0.84M$

Best regards.

7 0

3 years ago