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Dimas [21]
3 years ago
13

Name the SI base units that are important in chemistry, and give the SI units for expressing the following: (a)length, (b) volum

e, (c) mass, (d) time, (e) temperature.
Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer and Explanation:

The basic unit which are that are important in chemistry are meter, kilogram ,mol,m^3

Candela which is the unit of luminous of intensity is not so important in physics

(a) SI unit of length is meter (m)

(b) Si unit of volume is m^3

(c) Si unit of mass is kilogram (kg)

(d) SI unit of time is second (s)

(e) SI unit of temperature is kelvin (K)

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Answer: Microscope

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Provide an orbital energy level diagram for the geound state of a nitrogen atom.
Alenkasestr [34]

Answer:

A

Explanation:

Nitrogen is an atom made up of 7 electrons.

To draw the orbital energy level diagram, let us write the orbital notation of the atom;

  7 electrons of Nitrogen:

        1s² 2s² 2p³

So,

 The orbital notation diagram is :

            1s²         2s²          2p³

            ↑↓          ↑↓          ↑↑↑

6 0
3 years ago
What is the mass of an object with a density of 3.4 g/mL and a volume of 500.0 mL
umka2103 [35]
Formula: 
mass=density *volume

Given: 
Density=3.4 
Volume=500.0  

Plug them into the formula: 
mass=3.4 *500.0=1700

Final answer: 1700g
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Is neon element a metal nonmetal or metalloid
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3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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