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worty [1.4K]
4 years ago
7

What is the date of discovery of element Mercury

Chemistry
2 answers:
STALIN [3.7K]4 years ago
4 0

2000 BC and was found in tubes in Egyptian tombs dated from 1500BC

hope this helps

lesya692 [45]4 years ago
3 0
Before 2000BC by Chinese and egyptians
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An isotope of an element has the following symbol: 131 I How many electrons does it
WINSTONCH [101]

Answer:

D

Explanation:

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5 0
3 years ago
Determine the percent composition by mass of oxygen in the compound C6H1206
inysia [295]

Answer

% of oxygen(O) in glucose = 96/ 180 × 100 = 53.33

Explanation:

so the mass of oxygen in 1 mole of glucose = 6 × 16.00 g = 96.00 g. % oxygen = 96.00 × 100 % = 53.3 % by mass.

3 0
2 years ago
What quantity of energy would be necessary to boil water with a mass of
anygoal [31]

Answer:

i think 554 I think

Explanation:

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6 0
3 years ago
This is the chemical formula for acetone: CH32CO. Calculate the mass percent of carbon in acetone.
olga_2 [115]

Answer:

62.07 %

Explanation:

Chemical Formula =  (CH3)2CO = C3H6O

Mass of C = 12 g/mol

Mass of H = 1 g/mol

Mass of O = 12 g/mol

Mass of  C3H6O = 3(12) + 6(1) + 16 = 58 g/mol

Total mass of C in Acetone = 12 * 3 = 36 g

Mass Percent of C =  Total mass of C / Mass of Acetone   *  100

Mass percent = 36 / 58    *  100

Mass percent = 62.07 %

6 0
3 years ago
Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydroge
LuckyWell [14K]

Explanation:

The given data is as follows.

        Mass flow rate of mixture = 1368 kg/hr

      N_{2} in feed = 40 mole%

This means that H_{2} in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas (N_{2} + H_{2}) in feed stream.

Hence, calculate the total mass flow rate as follows.

           40 moles/hr of N_{2}/hr (28 g/mol of N_{2}) + 60 moles/hr of H_{2}/hr (2 g/mol of H_{2})

                  40 \times 28 g/hr + 60 \times 2 g/hr    

                  = 1120 g/hr + 120 g/hr

                  = 1240 g/hr

                  = \frac{1240}{1000}              (as 1 kg = 1000 g)

                  = 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

                 \frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr

                   = 110322.58 moles/hr

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of N_{2} into the reactor as follows.

                       0.4 \times 110322.58 moles/hr

                      = 44129.03 mol/hr

As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.

Therefore, calculate the rate flow of N_{2} into the reactor as follows.

                       0.028 kg \times 44129.03 mol/hr

                         = 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.

3 0
3 years ago
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