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olga_2 [115]
3 years ago
15

1. Why do we classify things?

Chemistry
1 answer:
mario62 [17]3 years ago
5 0

Answer:

to keep it organized in each groups

Explanation:

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What mass of Hg will occupy a volume of 75.0 mL?
k0ka [10]
75.0 mL in liters:

75.0 / 1000 => 0.075 L

1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L

moles Hg = 0.075 x 1 / 22.4

moles = 0.075 / 22.4

= 0.00334 moles of Hg

Hg => 200.59 u

1 mole Hg ----------------- 200.59 g
<span>0.00334 moles Hg ----- ( mass Hg )
</span>
mass Hg = 200.59 x 0.00334 / 1

mass Hg = 0.6699 / 1

= 0.6699 g of Hg




7 0
3 years ago
How many moles are in 22 grams of argon
Strike441 [17]
<span>number of moles are the amount of substance of an element .
number of moles can be calculated as follows;
number of moles = mass present / molar mass of element
molar mass of Ar - 40 g/mol
number of moles of Ar = 22 g / 40 g/mol = 0.55 mol .
there are 0.55 mol of Ar</span>
6 0
3 years ago
Read 2 more answers
Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin
Irina18 [472]
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ 
λ   = 0.693 / 32          (1) 

Nt = Nο eΛ(-λt)          (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),


Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt =  0.085Nο 

Percentage = (Nt/Nο) x 100%
                   = (0.085Nο/Nο) x 100%
                   = 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

8 0
3 years ago
A student plans to use a density versus solution concentration standard curve to identify the sodium chloride concentration in a
Art [367]

Answer:

a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g;

g. 0.1 g, 0.5 g, 1.0 g, 1.5 g, 2.0 g

Explanation:

The percent mass is defined as a ratio between the mass of a solute and mass of a solution:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%

Since solution only consists of a solute and solvent, express its mass as:

m_{solution} = m_{solute} + m_{solvent}

Then:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%=\frac{m_{solute}}{m_{solute} + m_{solvent}}

Firstly, solve for how much mass is required to prepare 3.0 %. Let's say, we have x g of the solute:

0.03 = \frac{x}{30.0 + x}\therefore x = 0.03(30.0 + x)

x = 0.90 + 0.03x

0.97x = 0.90\therefore x = 0.93 g

Similarly, solve for 6.0 %, let's say, we have x g of the solute again:

0.06 = \frac{x}{30.0 + x}\therefore x = 0.06(30.0 + x)

x = 1.80 + 0.06x

0.94x = 1.80\therefore x = 1.91 g

Hence, masses should be in a range of 0.93 g to 1.91 g.

7 0
3 years ago
How many grams are there in 3.3 x 1023 molecules of N2I6?
FromTheMoon [43]

Answer:

The number of grams is 434.5 g.

Explanation:

5 0
2 years ago
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