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3 years ago

12.A piece of magnesium is in the shape of a cylinder with a height of 5.62 cm

Chemistry

1 answer:

yaroslaw [1]3 years ago

4 0

Answer:

Density, d = 1.779 g/cm³

Explanation:

The density of a material is given by its mass per unit volume.

Here, height of a piece of magnesium cylinder, h = 5.62 cm

Its diameter, d = 1.34 cm

Radius = 0.67 cm

Volume of he cylinder,

$V=\pi r^2 h\\\\\text{Putting the value of r and h, we get :}\\\\V=(\pi \times (0.67)^2\times 5.62)\ cm^3$

$d=\dfrac{m}{V}\\\\d=\dfrac{14.1\ g}{(\pi \times (0.67)^2\times 5.62)\ cm^3}\\\\d=1.779\ g/cm^3$

So, the density of the sample is 1.779 g/cm³.

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Nonmetals gain electrons under certain conditions to attain a noble gas electron configuration. How many electrons must be gaine

gogolik [260]

Answer:

Explanation:

For non metals to attain a noble gas configuration, they gain the number of electrons needed to attain the noble gas configuration of the noble gas at the end of their periods. This means that these non metals would only take up the configuration of the last element on their periods which of course is always a noble gas.

The last element on the hydrogen period or more conservatively the only other element on the hydrogen period is helium, with an atomic number of 2. The atomic number is the number of protons in he nucleus of an atom. For an electrically neutral atom, the number of electrons equal the number of protons.

Hence we can deduce that helium has 2 electrons while hydrogen has one electron. Thus for it to attain the configuration of helium, it just needs to gain one more electron

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3 years ago

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

$2N_2O \rightarrow 2N_2 + O_2$

rate constant $k = 1.94\times 10^{-4} min^{-1}$

Half life $t_{1/2} = \frac{0.693}{k} = 3572 min$

Initial pressure $N_2 O = 4.70 atm$

Pressure after 3572 min = P

According to first order kinematics

$k = \frac{1}{t} ln\frac{4.70}{P}$

$1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}$

solving for P we get

P = 2.35 atm

$2N_2O \rightarrow 2N_2 + O_2$

initial 4.70 0 0

change -2x +2x +x

final 4.70 -2x 2x x

pressure of $O_2$ after first half life = 2.35 = 4.70 - 2x

x = 1.175

pressure of $N_2$ after first half life = 2x = 2(1.175) = 2.35 ATM

Total pressure = 2.35 + 2.35 + 1.175

= 5.875 atm

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3 years ago

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A resistor is placed in a circuit. The voltage of the circuit is changed and the current is measured. The graph seen here shows

Arturiano [62]

Answer- 2 Ohms

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2 years ago

How many particles are present in 12.47 grams of NaCl

liq [111]

Answer:

1.26*10²³ particles are present in 12.47 grams of NaCl

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, first of all you must know the amount of moles that represent 12.47 grams of NaCl. For that it is necessary to know the molar mass.

You know:

Na: 23 g/mole
Cl: 35.45 g/mole

So the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole

Now you apply a rule of three as follows: if 58.45 grams are present in 1 mole of NaCl, 12.47 grams in how many moles will they be?

$moles=\frac{12.47 grams*1 mole}{58.45 grams}$

moles= 0.21

You apply a rule of three again, knowing Avogadro's number: if in 1 mole of NaCl there are 6,023 * 10²³ particles, in 0.21 moles how many particles are there?

$number of particles=\frac{0.21 moles*6.023*10^{23} }{1 mole}$

number of particles= 1.26*10²³

1.26*10²³ particles are present in 12.47 grams of NaCl

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3 years ago

Which of the cells shown is a prokaryote?

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The answer is c because bacteria is prokaryote.

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3 years ago

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