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3 years ago

What happens to matter during chemical changes?

1 answer:

7 0

The object that goes through chemical change, changes completely to where you can not change it back to its original form. Physical change you can undo

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Which water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B prot

sergiy2304 [10]

Answer:

10 cm

Explanation:

water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B protection is 10 cm . the bar graph attached in the shows that the lengths of yellow and brown bars differ the most at 10 cm.

Hence,The biggest difference between the survival rates of UV-B protected and unprotected can be seen at the depth of 10 cm.

6 0

3 years ago

A 60-kg box is being pushed a distance of 6.2 m across the floor by a force P with arrow whose magnitude is 150 N. The force P w

Afina-wow [57]

Answer:

Explanation:

mass of box, m = 60 kg

distance, d = 6.2 m

force, P = 150 N

coefficient of kinetic friction, μk = 0.21

work done by the gravitational force, Wg = Force x distance x cos 90

Wg = 0 J

Work done by the normal force, Wn = force x distance x Cos 90

Wn = 0 J

Work done by the friction force, Wf = - μk mg d

Wf = - 0.21 x 60 x 9.8 x 6.2 = - 765.58 J

Work done by the applied force, Wp = P x d

Wp = 150 x 6.2 = 930 J

3 0

3 years ago

If the units of your answer are kg ∙ m2/s3, which of the following types of quantities could your answer be?

kirza4 [7]

Answer:

D. power

Explanation:

kg represents mass

(m/v)² represents velocity squared

Then kg·m²/s² represents mass·velocity² = kinetic energy or potential energy or work.

kg·m²/s³ will be the rate of doing work, which is power

4 0

3 years ago

Read 2 more answers

An atom of element X has one more shell of electrons than an atom of beryllium, but it has one less valence electron than beryll

3241004551 [841]

The correct answer is Sodium

8 0

3 years ago

Read 2 more answers

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago