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Alex17521 [72]
3 years ago
5

Fill in blanks plz help 23 points

Chemistry
1 answer:
shtirl [24]3 years ago
6 0
I'm not sure about the second one so sorry for that! But, the other three blanks are length, graduated cylinder, and matter (I believe its matter, I'm sorry if thats wrong)

Again, I'm sorry I couldn't answer one and accurately answer the other one but I hope this helps you!
You might be interested in
The force against a solid object moving through a gas or a liquid
statuscvo [17]

A force or a type of friction?

Force: Resistance--> If you go swimming, you feel the water pushing against you, making it harder to walk in water and even to swim unless you are doggy pattling or front strokes. Resistance is a force going against a solid object.

Friction: Fluid friction which is friction that occurs after a solid object travels through a liquid or gas

5 0
3 years ago
If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form
Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

Learn more about empirical formula : brainly.com/question/11588623

3 0
2 years ago
Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super
dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:X^{m+}Y^{n-}\rightarrow X_nY_m

Thus, for all the given combinations, we obtain:

- Y⁻

X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3

- Y²⁻

X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3

- Y³⁻

X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY

Best regards!

8 0
2 years ago
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

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

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 a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of
Likurg_2 [28]
The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer: 
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
                        (HBr)      (KOH)

Therefore, M2 = M1V1/V2
                        = 1.4 X 15.4/22.10
                        = 0.9756 M

Concentration of KOH is 0.9756 M.

3 0
3 years ago
Read 2 more answers
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