1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
rjkz [21]
3 years ago
15

Individual M has longer limbs and muscles than Individual N. Both individuals lift the same amount of weight from the ground to

their shoulders an equal number of times. If all repetitions were completed in synchrony, which statement about the amount of work completed by the individuals is true? At the end of the workout:
A. Individual M will have done less work because the object was moved a greater distance
B. both individuals will have done the same amount of work because force was applied for the same duration of time.
C. Individual M will have done more work because the muscle contracted faster to achieve synchrony with Individual N
D. Individual N would have done less work because the object was moved a shorter distance
Chemistry
1 answer:
Elena L [17]3 years ago
5 0

Answer:

D. Individual N would have done less work because the object was moved a shorter distance

Explanation:

When a body is moved a given distance (x) by a specific force (F), the work done on the body is equivalent to F*x. This shows that work is proportional to the applied force and the distance traveled by the body. Therefore, the higher the distance the higher the work done and the lower the distance the lower the work done. The work done by Individual N is lower due to the shorter limbs that correspond to shorter distance.

You might be interested in
1. Which of the following is NOT a cation?
Bezzdna [24]
Hey Madoudou


The correct answer is option B (sulfate)


The reason is because "Sulfate" has a negative sign.

In order for it to be a cation, it must have positive sign such as "iron(lll)ion


I hope this helps~
3 0
3 years ago
Answer questions based on the lab activity.
Alika [10]

Answer:

Conduction, Convection and Conduction

Explanation:

8 0
3 years ago
What is the ionic charge for the calcium ion in CaCO3?
Lunna [17]
<span>The ionic charge of Calcium (Ca) in calcium carbonate (CaCO3) is 2+. CaCO3 has a neutral ionic charge sin CO3 has a 2- charge.</span>
3 0
3 years ago
The density of ethanol is 0.788 g/mL. What is the mass of 157 mL of alcohol?
Lina20 [59]

Answer:

<h3>The answer is 123.72 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 157 mL

density = 0.788 g/mL

We have

mass = 0.788 × 157 = 123.716

We have the final answer as

<h3>123.72 g</h3>

Hope this helps you

6 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
Other questions:
  • When metals form ions, they tend to do so by what
    7·2 answers
  • 2. Which types of changes observe the law of
    8·1 answer
  • the density of gold is 19.3 g/cm the density of iron pyrite is 5.0 g/cm . if a nugget of iron pyrite and a nugget of gold each h
    15·1 answer
  • The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final
    14·1 answer
  • What is included in each square (element) on the Periodic Table? I'm looking for four things
    7·1 answer
  • Calculate the morality of a 35.4% (by mass) aqueous solution of phosphoric acid (H3PO4). Molar mass of H3PO4 is 98.00g/mol
    7·1 answer
  • I NEED HELP PLS HELP!
    12·1 answer
  • An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the
    10·1 answer
  • 0.385g of 17Cl is how many atoms?
    7·1 answer
  • When H2(g) reacts with F2(g) to form HF(g) , 542 kJ of energy are evolved for each mole of H2(g) that reacts. Write a balanced t
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!