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klasskru [66]
3 years ago
14

Select the correct answer. What is the SI unit of time?

Chemistry
2 answers:
photoshop1234 [79]3 years ago
7 0

Option A: second

SI units are international system of units. This is defined as a system of units of physical quantities such as mass, length, temperature, current etc. Some of the physical units are gram, meter, second etc. along with the set of prefixes which indicates the multiplication and division with some power of 10.

For same physical quantities, these units are inter convertible For example,

1 g=10^{-3}kg

and,

1 g=10^{3}mg

Here, g (gram), kg (kilogram) and mg (milligram) all are units of mass.

The units of time are second, minute, hour, day and year. Here, second is the SI unit and all the units are inter convertible as  follows:

1 year=365 days

1 day=24 hours

1 hour=60 minutes

1 minute-60 seconds

Thus, the SI unit of time is second.


Leni [432]3 years ago
5 0

Answer:

A. second

Explanation:

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Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many g of CO2 would be produced from the complete r
Alenkasestr [34]
<h2>1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3 </h2>

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}    

0.833M=\frac{\text{Moles of} HC_3H_3O_2\times 1000}{25ml}\\\\\text{Moles of} HC_3H_3O_2 =\frac{0.833mol/L\times 25}{1000}=0.0208mol

NaHCO_3+HC_2H_3O_2\rightarrow NaC_2H_3O_2+H_2O+CO_2

According to stoichiometry:

1 mole of HC_2H_3O_2 will give = 1 mole of CO_2

0.0208 moles of HC_2H_3O_2 will give =\frac{1}{1}\times 0.0208=0.0208 moles of CO_2

Mass of HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g

Thus 1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3

Learn more about molarity

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3 years ago
How many atoms are centered on the (100) plane for the fcc crystal structure?
Setler [38]

Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.

There will be 2 atoms in 100 plane of FCC crystal lattice.

In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.

For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.

Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 × \frac{1}{4} = 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.

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3 years ago
The following calculations must be handwritten in your notebook. – Acetic Acid ■ Hydrogen ion concentration ■ Ka ■ % Error – Ace
Nikitich [7]

Answer:

Explanation:

1) Acetic acid

Concentration is given as 0.103 M

The average pH of this solution = 2.96

we know that pH = - log [H+] therefore [H+] = 10-pH

[H+] = 10-2.96

= 1.1 x 10-3 M = 0.0011 M

Consider the equilibrium

CH3COOH ⇄CH3COO- + H+

Initial 0.103 0 0

Change -x +x +x

equlibrium 0.103 -x x x

Ka = x2 / 0.103 - x

Here the initial concentration of CH3COOH = 0.103 M

the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5

2) Acetic acid + NaOH

pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

Volume and conc of acetic acid = 10 mL of 0.103 M

= 10 mL x 0.103 mmol / mL

= 1.03 mmol

Volume and conc of NaOH added = 4 mL of 0.0992 M

= 4 x 0.0992 mmol

= 0.397 mmol

Consider the equation

CH3COOH + NaOH -----------> CH3COONa + H2O

Initial 1.03 0.397 0

Final 0.633 0 0.397

0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

pH = pKa + log 0.627 = pKa - 0.203

or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]

= 4.683

Ka = 10-4.683 = 2.07 x 10-5

3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

x =  1.03 x 10-4

Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99

(ii) Mass of Na2HPO4.7H2O =0.601 g

therefore no of moles = 0.601 / 268.07 = 0.00224 mole

= 2.24 mmol

The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml

= 0.075 M

consider the equilibrium

HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

(iii) Mass of Na3PO4.12H2O taken = 0.208 g

moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

Consider the equilibrium reaction

PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

equilibrium 0.055-x x x

Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]

x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

That is the conce of OH- ion = 0.025M

Therefore pH = 14 - pOH = 14 - 1.6 =12.4

3 0
3 years ago
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