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3 years ago

8

The number of dots around an element symbol in a Lewis dot structure equals the total number of electrons in an atom of that ele

ment.
True or False?

1 answer:

strojnjashka [21]3 years ago

8 0

The number of dots represents the amount of valence electrons, which is the same as the last digit of the elements group number in ptof.

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3 years ago

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3 years ago

How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr

pickupchik [31]

Answer: There are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = $87.5^{o}C = (87.5 + 273) K = 360.5 K$

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

$722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm$

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

$0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}$

Thus, we can conclude that there are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

5 0

3 years ago

The half-life of gold-198 is 2.7 days. After

Pie

Answer: 8.1 days

Explanation:

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

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= $\frac{x}{4}$

a) to find rate constant

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

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

$k=\frac{0.693}{2.7days}=0.257days^{-1}$

b) for completion of one fourth of reaction

$t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}$

$t=\frac{2.303}{0.257}\log{4}$

$t=8.1days$

Thus after 8.1 days , one fourth of original amount will remain.

8 0

3 years ago