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joja [24]
3 years ago
13

Please show the work :(What is the estimated force applied to the box if the acceleration is .40 m/s2?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
8 0

Answer:

20N

Explanation:

Given parameters:

        Force(N)     Acceleration(m/s²)  

            10                   0.2

           ?                      0.4

Unknown:

The force applied when the acceleration is 0.4m/s²

Solution:

From newton's second law of motion;

 Force = mass x acceleration

 Since we are using the same box, let us find the mass of the box;

      Force  = mass x acceleration

         10 = mass x 0.2

            mass = \frac{10}{0.2}  = 50kg

Now,

 The force in the second instance will be;

      Force  = 50 x 0.4  = 20N

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Answer:
            But-2-enoic acid has 11 Sigma Bonds and 2 Pi Bonds.

Explanation:
                   The sigma bonds which are formed due to head to head overlap of partally filled orbitals are shown in red color, while Pi bonds which are formed after the formation of sigma bond by overlap of orbitals perpendicular to the sigma bond are shown in blue color.

8 0
3 years ago
A gas has a pressure of 550 kPa at 327°C. What will its pressure be at 27°C, if the volume does not change?
lozanna [386]

Answer:

300°

Explanation:

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3 years ago
A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
BartSMP [9]

Answer:

Q_{metal} = -6799\,J

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

6 0
3 years ago
Read 2 more answers
PLEAZE HELP!!!
Alla [95]

Answer:

V= 12mL

Explanation:

you had the right idea with your Significant figures however, when we divide we see that it requires 2 significant figures as our least amount. this is because when looking at our division, 62 has 2 sig. fig. while 5.35 has a total 3. when looking at your answer we see that you had a total of 3 sig. figures. so in actuakity you had to round up to 12 and not to the tenths because the decimal makes .6 count as your third sig fig.

7 0
3 years ago
Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

  • For a weak acid like HF, the dissociation of HF will be:

<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>

[H₃O⁺] = [F⁻].

<em>∵ [H₃O⁺] = √Ka.C,</em>

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>

<em></em>

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>

6 0
3 years ago
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