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1 year ago

Need asap thank you

Chemistry

1 answer:

cupoosta [38]1 year ago

6 0

Answer:

c.boron-11

Explanation:

The atomic mass of boron is 10.81 u.

And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.

To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:

(10u)(x)+(11u)(1−x)100%=10.81u

Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.

Solving for x we get:

11u−ux=10.81u

0.19u=ux

x=0.19

1−x=0.81

And thus the abundance of boron-11 is roughly 81%.

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lord [1]

Answer:

Look at your periodic table

Explanation:

It tells you everything like Potasium 1 and 1 ox and 1 hydrogen

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2 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

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2 years ago

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2 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$