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Vaselesa [24]
3 years ago
12

A stock solution of HNO3 is prepared and found to contain 14.2 M of HNO3. If 25.0 mL of the stock solution is diluted to a final

volume of 0.500 L, the concentration of the diluted solution is ________ M.
Chemistry
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

First convert volume given into same unit.

Therefore 1000 mL =1L

1000mL=1L

25.0mL=?

(25.0×1)÷1000=0.025L

but using the equation;M1×V1=M2×V2

M1=14.2M

V1=0.025L

M2=?

V2=0.5L

Therefore;. 14.2×0.025=M2×0.5

M2=(14.2×0.025)÷0.5

M2=0.71M.

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The rock was white and black speckled with a density of 2.6 g/mL. Physical or chemical property?
stepladder [879]

Answer:

Chemical

Explanation:

7 0
3 years ago
Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th
Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ =  C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0
2 years ago
Which statement about Fe is supported by the modern atomic theory but not John Dalton’s theory?
lana66690 [7]

Answer:

The correct option is B

Explanation:

One of the claims of John Dalton's atomic theory is that atom is the smallest unit of matter (which suggests that there are no particles smaller than an atom in any matter). This claim has been disproved by the modern atomic theory which established that there are particles smaller than atom (called subatomic particles). These particles are electrons, protons and neutrons.

One of the modern atomic theory was by Neils Bohr, who proposed that <u>electrons move in circular orbits around the central nucleus</u>. Thus, the electrons of iron can also be said to be present in a region of space (circular path) around the nucleus. This proves that option B is the correct option as John Dalton's theory did not even recognize the electron(s) nor the nucleus.

5 0
2 years ago
Read 2 more answers
How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0
2 years ago
The order of components in a typical flame atomic absorption spectrometer is
Zinaida [17]

The order of components in a typical flame atomic absorption spectrometer is hollow cathode lamp--flame--monochromator--detector

<u>Explanation:</u>

  • The hollow cathode lamp practices a cathode created of the element of interest with a low internal pressure of inert gas.
  • Remove scattered light of other wavelengths from the flame. AAS flame includes aiming at first the fuel than the oxidant and then lighting the flame with the instrument's auto-ignition system. Applying flame Ddtroy any analyte ions and breakdown complexes.
  • The process  of the monochromator is to divide analytical lines photons moving through the flame
  • Photomultiplier tube (PMT)  as the detector the PMT determines the intensity of photons of the analytical line exiting the monochromator.
4 0
3 years ago
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