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zubka84 [21]
3 years ago
12

18. The mass of a quantity of NiCl2 is 24.6 g. How many formula units are in the sample?

Chemistry
2 answers:
MrRissso [65]3 years ago
8 0

its not A for sure becz i took the test i but b 1.9 3 1027 NiCl2 formula units

it was wrong answer

Charra [1.4K]3 years ago
4 0

Answer is: A. 1.1 3 1023 NiCl2 formula units.

m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.

M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.

n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).

n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.

n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.

Na = 6.022·10²³ 1/mol; Avogadro constant.

N(NiCl₂) = n(NiCl₂) · Na.

N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.

N(NiCl₂) = 1.13·10²³; number of formula units.

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Which observation provided Albert Einstein the clue that he needed to explain the photoelectric effect? Light is made up of extr
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The correct answer is

Energy of electrons depends on light’s frequency, not intensity.

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Unlike acceleration and velocity, speed does NOT need to specify
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C) mass.

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Speed analysis is divided into two main topics: average speed and instantaneous speed. It is considered a vector quantity, that is, it has a module (numerical value), a direction (Ex .: vertical, horizontal) and a direction (Ex .: forward, upwards). However, for elementary problems, where there is displacement in only one direction, the so-called one-dimensional movement, it is advisable to treat it as a scalar quantity (with only numerical value).

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3 0
3 years ago
The mass of an electron is about 9.11 × 10 -9 kg., write this whole number​
Pepsi [2]

Answer:0.00000009

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8 0
3 years ago
A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5
Citrus2011 [14]

Answer: 2948

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Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}years^{-1}

t = age of sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = \frac{70}{100}\times 100=70

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}

t=2948years

Thus the fossil is 2948 years old.

5 0
3 years ago
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