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Jet001 [13]
3 years ago
12

How many oxygen atoms are in 8.20 g Na 2SO 4?

Chemistry
1 answer:
dusya [7]3 years ago
3 0

Answer:

No. of atom =

no.of moles x avagardro's number xatomicity

= weight /molar mass x No x atomicity

=8.2/142 x6.02x10^23 x 4

=0.346 x 10^23(approximately)

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If the oxidation number of nitrogen in a certain molecule changes from +3 to -2 during a reaction, is the nitrogen oxidized or r
aliina [53]
The nitrogen has been reduced or has undergone reduction and it has gained one electron
7 0
3 years ago
A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen
Elenna [48]
Using the significant figure it would be 27.3
8 0
3 years ago
A device that does work with only one movement and changes the size or direction of a force is a(n) ____.
worty [1.4K]

Answer:

A device that does work with only one movement and changes the size or direction of a force is a simple machine.

Explanation:

  • For applying force, any used basic mechanical devices are simple machines.
  • Simple machine changes the direction as well as the amplitude of the applied force i.e. we can increase or decrease the magnitude of the force.
  • A simple machine is the most basic mechanism to use the force as we need in big mechanical machines.
  • Some of the examples of simple machines are inclined plane, lever, wedge, wheel and axle, pulley, and screw.
6 0
3 years ago
What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?
hjlf

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v =  1560 g / 41.2 g/ml

v = 37.9 ml

7 0
3 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
3 years ago
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