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Ahat [919]
2 years ago
5

The definition of the acceleration states that an object accelerates when it PLEASE HELP!!!

Chemistry
2 answers:
velikii [3]2 years ago
6 0

Maybe this can help.

In mechanics, speed increase is the pace of progress of the speed of an article regarding time (acceleration). Speed increases are vector amounts (in that they have greatness and direction). The direction of an item's speed increase is given by the direction of the net power following up on that article. The size of an item's speed increase, as depicted by Newton's Second Law, is the consolidated impact of two causes:

the net equilibrium of all outer powers acting onto that item — size is straightforwardly relative to this net coming about force;

that article's mass, contingent upon the materials out of which it is made — extent is conversely relative to the item's mass.

Rzqust [24]2 years ago
5 0

<u>There are two ways of thinking about it</u>:

  • First way: The definition of the acceleration states that an object accelerates when it <u>has a force applied to it</u>
  • Second way: The definition of the acceleration states that an object accelerates when it <u>changes its velocity</u>

<u></u>

<u><em>Reasons</em></u>

  • <u><em>For the first way</em></u><em>, by newton's second law, force equals to the product of mass times acceleration. The mass of the objects stays constant thus it is a constant in this equation. Thus when the Force isn't zero or bascially when it is applied, then the object must accelerate</em>
  • <u><em>For the second way</em></u><em>, acceleration is the rate of change of velocity, thus as velocity changes, the object must be accelerating</em>

<em />

Hope that helps!

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how many ml of 6.00 m hcl solution is needed to make 25.0 ml of 3.00 m hcl solution? answer in ml with correct significant figur
evablogger [386]

The answer is 12.5 mL.

Solution:

<u>M, 6.00M          V₁ =?</u>

M2 = 3.00M,    V₂ = 25.0mL

<u>M1V1 = M2 V2/M1</u>

V1 = M2V2 M,

V1 = 25.0X3.00/6.00

∴ V1 = 12.5 mL

The initial concentration of the resulting solution is 2 molar which is the volume multiplied by the concentration. The volume of the starting solution is therefore 2,500 ml × 2.25 mol. Calculate the number of moles of HCl dissolved in a given volume of acid by reacting an acid with CaCO3.

This is essentially a double exchange reaction with the decomposition of one of the products. To determine the amount of stock solution required divide the number of moles of glucose by the molarity of the stock solution.

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8 0
1 year ago
You have a hydrate of compound X that is 60.26g. You heat the hydrate, allow it to cool, and then remass it. The new mass is 23.
Alex17521 [72]

Answer:

100gmol

Explanation:

5 0
3 years ago
What is the temperature if 2L of 3moles HCl react with 2atm?
swat32

Answer:

T=16K=-256.8\°C

Explanation:

Hello!

In this case, according to the gas laws, we would be able to assume HCl can be modeled as an ideal gas for this calculation purpose; in such a way, we use the following equation to compute the temperature:

PV=nRT\\\\T=\frac{PV}{nR}

In such a way, we plug in moles, volume and pressure to obtain:

T=\frac{2atm*2L}{3mol*0.08206\frac{atm*L}{mol*K}}\\\\T=16K=-256.8\°C

Best regards!

8 0
3 years ago
Once water reach its boiling point what does you notice about its temperature??​
s2008m [1.1K]
It will stay at a constant temperature while it is boiling which is 212°F
6 0
3 years ago
A 0.568-g sample of fertilizer contained nitrogen as ammonium sulfate, . It was analyzed for nitrogen by heating with sodium hyd
Mariulka [41]

Answer:

6.69%

Explanation:

Given that:

Mass of  the fertilizer = 0.568 g

The mass of HCl used in titration (45.2 mL of 0.192 M)

= 0.192*\frac{45.2}{1000}* \frac{36.5}{1 \ mole}

= 0.313 g HCl

The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)

= \frac{44.3 \ mL * 1.0 \ L}{1000 \ mL} *0.133 \ mole/L

= 0.0058919 mole of NaOH

From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process

Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g

= 0.2151 g HCl

From above ; the total amount of HCl used = 0.313 g

The total amount that is used for complete neutralization = 0.2151 g

∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g

= 0.0979 g

We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g

Now; the amount of HCl neutralized by 0.0979 HCl = \frac{17}{36.5}*0.0979

= 0.0456 g

Therefore, the mass of nitrogen present in the fertilizer is:

= 0.0456 \ g \ NH_3 * \frac{1  \ mol \ NH_3 }{17.0 \ mol \ of \ NH_3} * \frac{1  \mol \ (NH_4)_2SO_4}{2 \ mol \ NH_3 } * \frac{2 \ mol \ N }{1  \mol \ (NH_4)_2SO_4}* \frac{14.0 g }{1 \ mol \ N}

= 0.038 g

∴ Mass percentage of Nitrogen in the fertilizer = \frac{0.038 \ g}{0.568 \ g} * 100%

= 6.69%

8 0
4 years ago
Read 2 more answers
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