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rjkz [21]
3 years ago
10

When 1 g of sodium chloride (NaCl) is placed in 100 g of water, a solution results. Once the solution is prepared, water is now

considered what part of the solution?
Chemistry
2 answers:
tiny-mole [99]3 years ago
7 0
Once the solution is prepared, water is now considered  as the solvent of the solution. On the other hand, NaCl is the solute. The solvent is a substance that dissolves other substances. It has a larger composition with respect to the other components in the solution.
Svetlanka [38]3 years ago
6 0

Answer: Solvent

Explanation:

Binary Solution is a homogeneous mixture of two components called as solute and solvent. Solute is the component which is present in smaller proportion and is solid for solid in liquid solution and solvent is the component which is present in larger proportion and is liquid for solid in liquid solution.

When 1 g of sodium chloride is dissolved in 100 g of water, it results into formation of binary solution where sodium chloride acts as solute and water acts as solvent.

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) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7
Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

2O_3\rightleftharpoons 3O_2

Rate of formation of oxygen : 7.78\times 10^{-1} M/s

Rate of the reaction(R) =\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}

R=\frac{1}{3}\frac{d[O_2]}{dt}

Rate of formation of oxygen=3 × (R)

7.78\times 10^{-1} M/s=3\times R

Rate of the reaction(R): 0.2593 M/s

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

R=-\frac{1}{2}\frac{d[O_3]}{dt}

\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s

-0.5186 M/s is the rate of the loss of ozone.

6 0
3 years ago
There are two different compounds of sulfur and fluorine.
olga nikolaevna [1]
What's your question......................
5 0
3 years ago
Hydrogen iodide is not produced by the same method as for hydrogen chloride.why??
lara [203]

Answer:

Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...

Using an ionic chloride and Phosphoric acid

H3PO4 + NaCl ==> HCl + NaH2PO4

H3PO4 + NaI ==> HI + NaH2PO4

H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.

2I-² === I2 + 2e-

Explanation:

4 0
3 years ago
A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?
Rama09 [41]

The frequency of the radiation is equal to 7.17  \times 10^{19} Hertz.

<u>Given the following data:</u>

  • Photon energy = 4.75 \times 10^{-14} Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

E = hf

<u>Where:</u>

  • h is Planck constant.
  • f is photon frequency.

Substituting the given parameters into the formula, we have;

4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}

Frequency, F = 7.17  \times 10^{19} Hertz

Read more: brainly.com/question/16901506

4 0
2 years ago
Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1
Nikitich [7]

Answer:

The answer is 3-Phenylpropanoic acid (see attached structure)

Explanation:

From spectral data:

3005 cm-1 ⇒ carboxylic acid (broad band)

1670 cm-1 ⇒ C=C

1603 cm-1 ⇒ Aromatic C-C bond

H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding  (clearer investigation of  spectrum would be expedient).

Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.

7 0
3 years ago
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