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ikadub [295]
3 years ago
14

Which career field is an applied science? Which hypothesis is testable?

Chemistry
1 answer:
VashaNatasha [74]3 years ago
4 0

Chemists, biologist, pharmacist, medical researcher ect…

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The primary gas in a volcano is: water vapor carbon dioxide sulfur dioxide nitrogen
Snezhnost [94]

Answer:

Water vapor

Explanation:

The magma consist of dissolved gases when these gases produce the force the volcanic eruption take place. The volcanic gases comes out and their volume is increased tremendously. The gases present in volcano are listed below:

The volcanic gases consist of water vapors, carbon dioxide and sulfur.

These three re the primary gases but the water is present in higher amount.

The percentage of water is 60%.

The carbon dioxide present in 10-40%.

Other gases present in valcano are nitrogen, argon, helium, neon methane and hydrogen.

7 0
3 years ago
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Which compound is insoluble in water?<br> (1) BaSO4 (3) KClO3<br> (2) CaCrO4 (4) Na2S
pav-90 [236]
 <span>BaSO4 
I hope this helps!</span>
7 0
3 years ago
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What acid and what base would you choose to prepare the salt potassium perchlorate (kclo4)?
Nonamiya [84]
HClO₄ + KOH → KClO₄ + H₂O

HClO₄ - perchloric acid
KOH - potassium hydroxide
8 0
3 years ago
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Name gaseous element in period 2 group 16?
Sholpan [36]

Answer:

mark me as a brainlist answer

Explanation:

Oxygen

3 0
3 years ago
A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r
Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added =  0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 ×  0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 =  0.005 moles -  0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3  in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0
2 years ago
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