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Greeley [361]
2 years ago
14

Why is it necessary to specify temperature when giving Ka value​

Chemistry
1 answer:
Yuliya22 [10]2 years ago
5 0

Answer:

Because the equilibrium constant is only at constant at temperature. Changing the temperature changes the Ka.

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What is the scaling factor if the molar mass of
Step2247 [10]

Answer:

The scaling factor is 5.

Explanation:

Hello there!

In this case, since the scaling factor is defined as the ratio of the molar mass of the molecular formula (complete) to the empirical formula (simplified), it is possible to compute it for the empirical formula of CH2O whose molar mass is 30 g/mol (12+2+16) as shown below:

sf=150/30=5

Therefore, we can also infer that the molecular formula would be:

C_5H_1_0O_5

Best regards!

6 0
3 years ago
A plant release it’s seed , but soon there after, the area seeds are in experiences a yearlong drought. How would the seeds most
SOVA2 [1]

Answer:

Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.

Explanation:

Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.

6 0
3 years ago
Anyone plz help?? T^T
Wewaii [24]

Answer:

C. Mg3N2

Explanation:

--------------

4 0
3 years ago
Read 2 more answers
Were would you be if all the stars you see are circumpolar
8_murik_8 [283]
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3 0
3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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