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telo118 [61]
3 years ago
14

Etahnol has a density of 0.79 g/ml. what is the volume, in quarts, of 1.80 kg of alcohol?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
7 0
Density =mass /volume

Volume =1800÷0.79
= 2278.481ml
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ArbitrLikvidat [17]

Answer:

Option B

Explanation:

Magnesium is an alkali earth metal and chlorine is a nonmetal. Due to this, and their differences in electronegativity, the elements form an ionic bond. Since magnesium has two valence electrons, it will give them up to two chlorine atoms (which has 7 valence electrons), which needs them to complete its octet. From this, magnesium forms a cation, and will have the ion formula Mg(2+). The two chlorine atoms form anions, and will have the ion formula Cl(1-).

Terms:

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7 0
2 years ago
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Consider the set of isoelectronic atoms and ions a2–, b–, c, d+, and e2+. which arrangement of relative radii is correct?
faust18 [17]
The term isoelectronic atoms means that if the given atoms are neutral, they would have the same number of electrons, which is relative to their sizes. Ions with positive charges are called cations, losing electrons. On the other hand, ions with negative charges are called anions, gaining electrons. The more electrons that the atoms have, the bigger it is in size. Hence, the arrangement of the radii of the atoms would have to be,

      e²⁺, d⁺, c , b⁻, a⁻

The arrangement is from smallest to largest.
7 0
3 years ago
Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate
nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

N=N_A\times n

Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

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3 0
3 years ago
calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure
andriy [413]

Answer:

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Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

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