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
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
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Answer:
See explanation.
Explanation:
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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:
Thus, for all the given combinations, we obtain:
- Y⁻

- Y²⁻

- Y³⁻

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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
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The question is incomplete. Complete question is attached below:
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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.