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4 years ago

Order the levels of organization from the smallest (1) to the largest (4).

Chemistry

1 answer:

yKpoI14uk [10]4 years ago

3 0

Answer:

cells=1

organ=3

organ system=4

tissue=2

Explanation:

i just did the question right now and i got them all correct.

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50 Points!! 25 Each Points Please Help ASAP!! View Attached Image!! Will Mark Brainliest If All Are Answered!!

Tema [17]

Answer:

Match the words to the definitions.

Explanation:

1. F

2. A

3. C

4. E

5. B

6. D

7. B

8. F

9. A

10. E

11. D

12. C

13. B

4 0

3 years ago

Which phrase is the best description of weathering?

Anit [1.1K]

I just had this question it is c

8 0

3 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)

Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

$pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414$

8 0

1 year ago

A recipe calls for 1.2 cups of oil. How many liters of oil is this?

Nadusha1986 [10]

.3 liters... im pretty sure this is correct!!

3 0

3 years ago

Read 2 more answers