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Paraphin [41]
3 years ago
5

Which of the four cars had the least net force applied to it when it was launched down the ramp

Chemistry
1 answer:
castortr0y [4]3 years ago
5 0

The least net force applied : Car 3(12 N)

<h3>Further explanation  </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

Car 1 ⇒m=0.5 kg, a=36 m/s²

\tt \sum F=0.5\times 36=18~N

Car 2⇒m=0.8 kg, a=50 m/s²

\tt \sum F=0.8\times 50=40~N

Car 3⇒m=0.6, a=20 m/s²

\tt \sum F=0.6\times 20=12~N

Car 4⇒m=1, a=19~m/s²

\tt \sum=1\times 19=19~N

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If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?
Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

  • Volume of Hexane dispensed = 40ml
  • Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

brainly.com/question/4721701

7 0
2 years ago
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr
lys-0071 [83]

Answer:

a. No change.    

b. The equilibrium will shift to the right.

c. No change

d. No change

e.  The equilibrium will shift to the left

f.  The equilibrium will shift to the right      

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining   Keq  again.

The equilibrium constant  for  A(s)⇌B(g)+C(g)  

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use  Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a.  double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´     ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp  ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to  double the pressures of A and B. Therefore there is no change.

b.  double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change  reduced the pressures by a half :

Q =  pB´x pC´  = (  1/2 pB ) x ( 1/2 pC )  =  1/4 pB x pC  = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the  concentration of B and halve the concentration of C

Doubling the concentrantion doubles  the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q =  pB´x pC´  = ( 2 pB ) x ( 1/2 pC ) = K

e.  double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q =  pB´x pC´   = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f.  double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q =  pB´x pC´   = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16  Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

3 0
3 years ago
The shells further away from the nucleus are larger/smaller and can hold more/less electrons.
Sergeu [11.5K]
The shells further away from the nucleus are LARGER and can hold MORE electrons
8 0
3 years ago
PLEASE HELP ASAP !!
Ivenika [448]

Answer:

  • 2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄
  • Theoretical yield of H₂SO₄ = 213 g
  • percent yield of H₂SO₄ = 94 %  

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

                       2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

                        2SO₂ +  O₂ + 2H₂O  -----------> 2H₂SO₄

                        2 mol                                         2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

                 volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

                 volume of gas = 2 mol x 22.4 L/mol

                 volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

            mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

        mass in grams = 2 mol x 98 g/mol

        mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

               44.8 L of SO₂ ≅ 196 g of H₂SO₄

               48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

                g of H₂SO₄  = 196 g x 48.6 L / 44.8 L

                g of H₂SO₄  =  213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

             percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

           percent yield = 200 g/ 213 g x 100 %

          percent yield = 94 %    

So percent yield of H₂SO₄ = 94 %    

8 0
3 years ago
Determine the mass of a gold bar that has a density of 19.3 g/cm3 and is 4.72 cm high by 8.21 cm long by 3.98 cm deep.
TEA [102]
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
5 0
3 years ago
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