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2 years ago

If an object is moving right at 10 N and a force is applied from the left at 8N, what is the net force on the object?

Chemistry

1 answer:

ryzh [129]2 years ago

5 0

Answer:

If an object is moving at a constant speed in a constant rightward direction, then the acceleration is zero and the net force must be zero.

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Ymorist [56]

Answer:

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Explanation:

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3 years ago

WILL GiVe bRaInLiEsT, 5 sTaR, AnD tHaNkS!!!!

Nataliya [291]

The answer is more protons than electrons.

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3 years ago

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago

Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

8 0

3 years ago

What is the concentration of the acid (H2SO4) in the following titration?

lapo4ka [179]

If your options are among the following:
<span>0.64 M 1.0 M 0.32 M 0.16 M.
Then the correct answer is 0.64 M. I hope this is what you were looking for</span>

3 0

2 years ago

Read 2 more answers