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3 years ago

If an electrical current can run though a solution, it is said to contain what?

Chemistry

2 answers:

lesantik [10]3 years ago

8 0

the answer i believe is electrolytes.

Nezavi [6.7K]3 years ago

5 0

Answer:

electrolytes

Explanation:

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1.-A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) +

Alika [10]

The number of mole of aluminum chloride, AlCl₃ produced when 5 moles of chlorine gas, Cl₂ reacts is 3.33 moles

<h3>Balanced equation </h3>

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

From the balanced equation above,

3 moles of Cl₂ reacted to produce 2 moles of AlCl₃.

<h3>How to determine the moles of AlCl₃ produced </h3>

From the balanced equation above,

3 moles of Cl₂ reacted to produce 2 moles of AlCl₃.

Therefore,

5 moles of Cl₂ will react to = (5 × 2) / 3 = 3.33 moles of AlCl₃.

Thus, 3.33 moles of AlCl₃ were obtained from the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

5 0

2 years ago

What are the two ways we can get the mass of an object?

Anni [7]

Explanation:

divide the objects weight by the acceleration of gravity to find the mass

7 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

What element is used in light bulbs as a filament???? help rly need help

frutty [35]

The answer would just be:
<span>Tungsten
</span>hope that this helps you! =)

6 0

3 years ago

Which compounds are lewis acids? select all that apply. (ch3ch2)2hc+ cl3c− ch3nh2 bbr3?

alexandr402 [8]

The Lewis acids are electron pair acceptors due to the availability of vacant orbitals they accept electrons from the donor compounds.

$(CH_3CH_2)_2CH^{+}$ is a Lewis acid due to the unfilled valence shell on carbon atom (possessing positive charge) it accepts electron readily.
$CCl_{3}^{-}$ is not a Lewis acid due to the presence of negative charge which represents that there are electrons available for donation.
$CH_3NH_2$ is not a Lewis acid due to the presence of lone pair on nitrogen which is easily donated to the acceptors.
$BBr_3$ is a Lewis acid due to the availability of vacant p-orbital on boron atom.

Hence, $(CH_3CH_2)_2CH^{+}$ and $BBr_3$ are Lewis acids.

3 0

3 years ago