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3 years ago

13

How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through

a solution containing CrCl3

1 answer:

True [87]3 years ago

5 0

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

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Which of the data sets qualitative data?

Kaylis [27]

Answer:

Uh, the second one?

Explanation:

Try to restate the question please.

3 0

3 years ago

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As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is

chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0

3 years ago

Pls help and i will mark brainliest

myrzilka [38]

Answer:

No. 3 lithium

Explain that lithium has 3 protons and 3 electrons. There are 2 electrons on the first energy level and 1 electron on the second

No. 4 Hydrogen (H) and helium (He) have a valence shell containing one and two electrons respectively. They make up the first period (row) of the periodic table. Their valence electron/s are in the first energy level (n=1) , as is denoted by 1s1 and 1s2 .

<em>That</em><em>'</em><em>s</em><em> </em><em>all</em><em> </em><em>i</em><em> </em><em>dont</em><em> </em><em>know</em><em> </em><em>if</em><em> </em><em>this</em><em> </em><em>is</em><em> </em><em>right</em><em> </em><em>though</em><em>.</em><em> </em>

6 0

3 years ago

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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

<u>Answer:</u> The equilibrium concentration of $COF_2$ is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

<u>Initial:</u> 2.00

<u>At eqllm:</u> 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

4 0

3 years ago

Which sample would serve as a better buffer and therefore a better environment for aquatic life?

Ann [662]

Answer:

sample A

Explanation:

the first one because of the ppm value

6 0

2 years ago

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