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algol13
3 years ago
15

How many watts of power (W) do you expend when you exert a force of 10 N that moves a

Chemistry
1 answer:
mihalych1998 [28]3 years ago
7 0

Answer:

Power = 7.5 watt

Explanation:

Given data:

Power expend = ?

Force applied = 10 N

Distance cover = 1.5 m

Time = 2 s

Solution:

Power = work/ time

First of all we will calculate work.

Work = Force × distance

Work = 10 N × 1.5 m

Work = 15 N.m

Now we will calculate the power.

Power = 15 N.m / 2s

N.m/s = 1 watt

Power = 7.5 watt

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How many atoms are in 45.22 moles of strontium​
AleksandrR [38]

Answer:

272.31× 10²³ atom of strontium

Explanation:

Given data:

Number of moles of strontium = 45.22 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

45.22 mol × 6.022 × 10²³ atom / 1 mol

272.31× 10²³ atom

7 0
3 years ago
at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were
Nataly [62]

<u>Answer:</u> The percent change in volume will be 25 %

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1

Putting values in above equation, we get:

\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L

Percent change of volume = \frac{\text{Change in volume}}{\text{Initial volume}}\times 100

Percent change of volume = \frac{(2-1.5)}{2}\times 100=25\%

Hence, the percent change in volume will be 25 %

5 0
3 years ago
1. I am in group two period three. What element am I?<br> A Ti<br> B, C<br> C. Ca<br> D. WE
AysviL [449]

Answer:

Ca

Explanation:

5 0
3 years ago
Read 2 more answers
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
Which of these conjugate acid-base pairs will not function as a buffer?A) HNO3 and NO3^-B) HCO3^- and CO3^2-C) C2H5COOH and C2H5
marin [14]
<h3><u>Answer;</u></h3>

A) HNO3 and NO3^-

<h3><u>Explanation;</u></h3>
  • <em><u>HNO3 is a strong  acid and NO3 is its conjugate base, meaning it will not have any tendency to withdraw H+ from solution.</u></em>
  • Buffers are often prepared by mixing a weak acid or base with a salt of that weak acid or base.
  • The buffers resist changes in pH since they contain acids to neutralize OH- and a base to neutralize H+. Acid and base can not consume each other in neutralization reaction.
5 0
3 years ago
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