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ValentinkaMS [17]
3 years ago
7

What is the volume of 90.0 g of ether if the density of the ether is 0.70 g/mL?

Chemistry
1 answer:
natali 33 [55]3 years ago
5 0
Mass = 90.0 g

Density = 0.70 g/mL

Volume = in mL ?

D = m / V

0.70 = 90.0 / V

V = 90.0 / 0.70

V = 128.57 mL
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Answer:

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Changing the number of ___ gives an atom a neutral charge or causes it to take on a positive or negative charge. (pick on that b
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Which of the following species has the greatest
Licemer1 [7]

Answer: Option (3) is the correct answer.

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4 0
4 years ago
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50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

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

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

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VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
In this double replacement reaction, which are the products? hcl + naoh —> ______?
Anarel [89]
The answer is:  " NaCl + H₂O " ; (or; write as:  " H₂O + NaCl " ) .
________________________________________________________
Specifically:
_________________________________________________________

    HCl + NaOH  —>  NaCl + H₂O  ;   or; write as:

    NaOH + HCl   —>  H₂O + NaCl  .
_______________________________________________________
            This type of "double-replacement" reaction is called "neutralization".
 
             Since we are adding a strong acid to a strong base (reactants), we know that the product will be:  1) a salt ; and 2) water.  Since we know one of the reactants will be "water" (H₂O) ; we can find the base (i.e. , the "remaining product") from selecting the "unused elements" to form the corresponding "salt".
________________________________________________________
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