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Katarina [22]
3 years ago
13

PLS I M IN HURRY.

Chemistry
2 answers:
ludmilkaskok [199]3 years ago
7 0

What is your question?

morpeh [17]3 years ago
6 0

Answer:

Use p1=560mmHg, v1=50cc

p2=760mmHg, then v2=p1v1/p2=36.84cc

Hope it helps

Explanation:

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Answer:

liquid

a semi permeable membrane

oxygen

Explanation:

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3 years ago
Which of the following best defines velocity?
Lina20 [59]

Answer: D

Explanation:

Because velocity is speed

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Which of the following is a common property of both strong acid and strong base?
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Answer:please can you help me with the last question that I asked I will help you if you help me please help

Explanation:

3 0
3 years ago
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If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
Octane has a density of 0.692 g/ml at 20∘c. how many grams of o2 are required to burn 17.0 gal of c8h18
german

156251.099 grams of O₂ are required to burn 17.0 gal of C₈H₁₈

<h3>Further explanation</h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g / cm³ or kg / m³

Density formula:

\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}

ρ = density

m = mass

v = volume

1 gal equal to = 3785.41 ml

then 17.0 gal = 17 x 3785.41 = 64351.97 ml Octane

grams Octane = ρ x ml

grams Octane = 0.692 g.ml x 64351.97

grams Octane = 44531.563

molar mass Octane (C₈H₁₈) = 114

mole Octane = grams : molar mass

mole Octane = 44531.563  : 114

mole Octane = 390.627

From the reaction

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9H₂O

mole C₈H₁₈ : mole O₂ = 1 : 25/2

mole\:O_2\:=\:\frac{25}{2} \times\:390.627

mole O₂ = 4882.846

grams O₂ = mole x molar mass

grams O₂ = 4882.846 x 32

grams O₂ = 156251.099

<h3>Learn more </h3>

moles of water

brainly.com/question/1405182

grams of oxygen required

brainly.com/question/8175791

mass of copper required

brainly.com/question/1680090

Keywords: Octane,  mole, mass, gal, density

5 0
3 years ago
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