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3 years ago

True or False. A force is a push or pull exerted on an object.

Chemistry

2 answers:

tangare [24]3 years ago

6 0

Answer:

True.

Explanation:

drek231 [11]3 years ago

4 0

Answer:

TRUE

Explanation:

remember newtons laws of motion. Every action has an equal and opposite reaction. LOL

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7. Explain how magnetic potential energy can be transformed into kinetic energy. (1<br> point)

Arisa [49]

The potential energy by the magnetic field can turn into kinetic energy once the field is moving from the S pole to the N pole when it reaches the N pole it is potential energy when it exits the S pole it is kinetic energy.

5 0

1 year ago

Help please this is also science idk why brainly doesn't have that option

solniwko [45]

Answer:

fffffffffffdddddddddddddddddd

Explanation:

fdddddddddddddddddddddddddd

6 0

2 years ago

Read 2 more answers

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,

White raven [17]

Explanation:

(a) potassium oxide with water

$K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)$

According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

$P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)$

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

$Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)$

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

$SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)$

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

6 0

3 years ago

What was the result of heating the mixture? All BUT ONE choice is correct.

snow_tiger [21]

Answer:

w gang alright

Explanation:

ay its b alright

4 0

3 years ago