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3 years ago

10

Which statement(s) is/are true? Energy

1 answer:

Anna007 [38]3 years ago

3 0

Answer:

D

Explanation:

it's not c because energy can't disappear or be destroyed so that makes E wrong

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What element is CU?plz helpppp

Ulleksa [173]

CU is the element symbol for Copper.

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3 years ago

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Calculate the kinetic energy of a 150 g baseball moving at a speed of 40. m/s (89 mph).

LUCKY_DIMON [66]

Kinetic Energy= 0.5 times velocity squared times mass
So KE= 0.5times 150 times 40 squared

(Use a calculator)
And don't forget units

6 0

3 years ago

Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.

Harlamova29_29 [7]

Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+<span>]/[HOCl] ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>) ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14

</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these

6 0

3 years ago

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Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

7 0

3 years ago

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i

alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

3 0

3 years ago