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aleksley [76]
3 years ago
7

???????????hahshshhshshsh

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

Answer:

structure

Explanation:

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PLEASE HELP ME. REAL ANSWERS ONLY PLZ
eimsori [14]

Answer:

Please make it full I'm not seeing it

8 0
3 years ago
How much concentrated solution would it take to prepare 2.75 L of 0.400 M HCl upon dilution with water
grin007 [14]
Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20g/mL. Calculate the molarity of the concd HCl.
1.20 g/mL x 1000 mL x 0.37 x (1/36.5) = about 12 M or so but you do it exactly.

Then mL x M = mL x M
mL x 12 M = 2800 mL x 0.475
Solve for mL of the concd HCl solution.
7 0
2 years ago
cyclohexanol is burned as a fuel, according to the following equation, how many moles of oxygen are needed to produce 13.7 mol o
lidiya [134]

 The moles  of oxygen  that are needed  to produce 13.7 moles  of  carbon  dioxide  is  21.17  moles  of Oxygen


<u><em>calculation</em></u>

2 C₆H₁₂O  + 17 O₂ → 12 CO₂  +12 H₂O


The moles of O₂  is determined using the mole ratio

that  is for  given equation above   O₂ : Co₂  is 17 :12

therefore the moles of O ₂= 13.7   moles x 17/12 =21.17  moles

3 0
3 years ago
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0
3 years ago
Which metal atom below would Not be involved in formation of a Type II Compound?
BartSMP [9]
In formation of a Type II Binary Compound, the metal atom present is<span> NOT</span> found in either Group 1 or Group 2 on the periodic table.  For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
4 0
3 years ago
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