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Korvikt [17]
3 years ago
11

How many significant figures are in each of the following measurements? (c) 0.03003 kg (f) 3.8200 x 103 L (a) (d) 0.00450 m 35.0

445 g (b) 59.0001 cm (e) 67,000 m2
Chemistry
1 answer:
erica [24]3 years ago
8 0

Answer:

c =four significant figures = 3003

f = Four significant figures = 3820

d =(i) three significant figures = 450

d =(ii) Six significant figures  = 35.0445

b = significant figures = 59.0001

e = five significant figures = 67000

Explanation:

c = 0.03003 kg

four significant figures = 3003

f = 3.8200 × 10³ L

Four significant figures = 3820

d =(i) 0.00450 m and (ii) 35.0445 g

(i) three significant figures = 450

(ii) Six significant figures  = 35.0445

b = 59.0001 cm

six significant figures = 59.0001

e = 67000

five significant figures = 67000

The given measurement have five significant figures 41006. this measurement can be rounded to three significant figures i.e, 0.410 g.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

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Describe the sources of water pollution in hydrological cycle.
Bingel [31]

Explanation:

The hydrological cycle is the continuous cycling of water between land, open water surfaces and the sea. This cycle begins with evaporation, sunlight evaporates water from the surface of earth, next condensation happens, the water absorbed is now used to form clouds, after these clouds are filled to the maximum, precipitation happens, this can be in the form of rainfall and snow, this cycle finalizes when the precipitation of water runs off the land and back into water sources.

Sources of water pollution:

  • <em>During precipitation: </em>Smog can be gathered in the atmosphere, during precipitation this pollution can turn into acid rain.
  • <em>During runoff:</em> After acid rain hits the ground this polluted water can run into water sources (lakes, rivers, reservoirs).To some extent rivers are a self-renewing resource, if a small quantity of pollution discharges in it the river can return to a clean, unpolluted condition, unfortunately, if the pollution is too big the renewing won't be possible, another problem is even though rivers get cleaned the pollution moves to the seas. Lakes are even more vulnerable to pollution, the flushing effect in these water bodies is less evident than in rivers.

I hope you find this information useful and interesting! Good luck!

8 0
3 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
Maurinko [17]

Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

7 0
3 years ago
What causes the lines in the spectrum for elements? a continuous release of energy from the heated element a quantum absorption
solmaris [256]

Answer:a quantum absorption of energy

Explanation:

Bohr’s model explains the spectral lines .While the electron of the atom remains in the ground state, its energy is unchanged. When the atom absorbs one or more quanta of energy, the electron moves from the ground state orbit to an excited state and when the atom relaxes back to a lower energy state, it releases energy that is again equal to the difference in energy of the two orbits.

5 0
2 years ago
Read 2 more answers
27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a
jok3333 [9.3K]

Answer : The volume of the cube is, 21.88cm^3

Solution : Given,

Density of nickel = 8.91g/cm^3

Number of nickel atoms = 2\times 10^{24}

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, 6.022\times 10^{23} atoms form 1 mole of nickel

So, 2\times 10^{24} atoms form \frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321 moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}

\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

Density=\frac{Mass}{Volume}

8.91g/cm^3=\frac{194.94g}{Volume}

Volume=21.88cm^3

Therefore, the volume of the cube is, 21.88cm^3

4 0
3 years ago
What is the concentration of hydronium ion in a 0.121 M HCl solution?
kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M              .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0
3 years ago
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