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iris [78.8K]
3 years ago
12

Which solution, solution X of pH 9.4 or solution Y of pH 7.5, is more acidic? Which solution, solution of 9.4 or solution of 7.5

, is more acidic? Solution X is more acidic. Solution Y is more acidic.
Chemistry
1 answer:
zzz [600]3 years ago
8 0

Answer:

Solution Y is more acidic

Explanation:

The pH value is an indicator of the concentration of hydrogen ions (H⁺) in an aqueous solution, and it is defined as:

pH= -log [H⁺]

⇒ [H^{+} ]= 10^{-pH}

Where  [H⁺] is the concentration of H⁺. A solution with a higher concentration of H⁺ is more acidic, and a solution with a lower concentration of H⁺ is less acidic or basic.

As result, <em>the lower the pH, the more the concentration of hydrogen ions ([H⁺]), and the aqueous solution is more acidic</em>.

Solution Y has a lower pH value (7.5) than solution X (9.4). So, solution Y is more acidic.

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CAN SOMEONE PLEASE HELP ME ASAP PLEASEE ANYBODY LITERALLY ANYONE OUT THERE PLEASE
IgorC [24]

Answer:

Electrons

Explanation:

Atoms can combine together and share electrons between them. Atoms that share electrons are linked to each other in a form, called a molecule that is in a lower energy state than either of the separate atoms alone.

Pls Brainliest! It would mean a lot! ;)

4 0
3 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act
algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y =  -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³  and  not 2.228 x 10⁻³). You can  recognize this mistake if you plot the data and notice it produces an outlier.

5 0
3 years ago
0.05 grams od cobalt ii oxalate will dissolve in one liter of solution at 25 degree celcius. Calculate the ksp for cobalt ii oxa
Veseljchak [2.6K]

Answer:

The equilibrium expression is:

CoC2O4(s)⇌Co2+(aq)+C2O2−4(aq)

For this reaction:

Ksp = [Co2+][C2O2−4]=1.96×10−8

Explanation:

Batteries will not clot if cobalt ions are removed from its cells. Some blood collection tubes contain salts of the oxalate ion,

C2O2−4

, for this purpose. At sufficiently high concentrations, the calcium

and oxalate ions form solid, CoC2O4·H2O (which also contains water bound in the solid). The concentration of Co2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of

C2O2−4

ion must be established before CoC2O4·H2O begins to precipitate.

CoC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CoC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Co2+], we can solve for the concentration of

C2O2−4

that is necessary to produce the first trace of solid:

5 0
3 years ago
In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH4 (g) 2O2
Degger [83]

Answer:

The energy released will be -94.56 kJ or -94.6 kJ.

Explanation:

The molar mass of methane is 16g/mol

The given reaction is:

CH_{4}(g) + 2O_{2} (g) --> CO_{2} (g)+ 2H_{2}O(l)

the enthalpy of reaction is given as  ΔH = -890.0 kJ

This means that when one mole of methane undergoes combustion it gives this much of energy.

Now as given that the amount of methane combusted = 1.70g

The energy released will be:

=\frac{energy released by one moleXgiven mass}{molarmass} =\frac{-890X1.7}{16}= -94.56 kJ

7 0
3 years ago
Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given
zimovet [89]
The generalized rate expression may be written as:
r = k[A]ᵃ[B]ᵇ

We may determine the order with respect to B by observing the change in rate when the concentration of B is changed. This can be done by comparing the first two runs of the experiment, where the concentration of A is constant but the concentration of B is doubled. Upon doubling the concentration of B, we see that the rate also doubles. Therefore, the order with respect to concentration of B is 1.
The same can be done to determine the concentration with respect to A. The rate increases 4 times between the second and third trial in which the concentration of B is constant, but that of A is doubled. We find that the order with respect to is 2. The rate expression is:

r = k[A]²[B]
4 0
3 years ago
Read 2 more answers
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