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Ganezh [65]
3 years ago
11

15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon

is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Chemistry
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

well, first off. the formula for carbon tetrachloride is CCl4

We need to find the molar mass of carbon tetrachloride

1(Mass of C) + 4(mass of chlorine)

1(12) + 4(35.5)

12 + 142

154 g/mol

Number of moles of CCl3 in 543.2g CCl3

n = given mass / molar mass

n = 543.2/153

n = 3.53 moles

always remember to brainly the questions you find helpful

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What does the following energy diagram represent?A. Activation energyB. Exothermic reactionC. Specific heat capacityD. Endotherm
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8 0
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3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)
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Explanation:

(a)   The given reaction equation is as follows.

        Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g) (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}

Reduction-half-reaction: NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

      Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l) (acidic)

(b)   The given reaction equation is as follows.

        HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq) (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)

Oxidation-half reaction: Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

      4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}

Therefore, balancing the whole reaction equation in the basic medium as follows.

      H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)

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