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Rama09 [41]
1 year ago
15

A sample of hydraulic acid has a volume of 2.3L. What is the volume of acid in gallons?

Chemistry
2 answers:
Igoryamba1 year ago
8 0

If a sample of hydraulic acid has a volume of 2.3L. then its volume of acid would be 0.5059 gallons.

<h3 /><h3>What is a unit of measurement?</h3>

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

as given in the problem, A sample of hydraulic acid has a volume of 2.3L

4.54609 L = 1 gallon

1 L = 1/4.54609 gallons

2.3 L = 2.3 /4.54609 gallons

        =  0.5059 gallons  

Thus, the volume of acid in gallons would be 0.5059 gallons.  

Learn more about the unit of measurement from here

brainly.com/question/12629581

#SPJ1

Assoli18 [71]1 year ago
7 0

0.60759572 gal  is the volume of acid in gallons  in a sample of hydraulic acid .

<h3>What do mean by the term "gallons" ?</h3>

The gallon is a unit of measurement for volume and fluid capacity in both the US  units and the British imperial systems of measurement.

In SI base units, 1 Gal is equal to 0.01 m/s²

Since ,

1 L, l = 0.2641720524 gal

now, volume of acid in gallons is given by -

2.3 L, l = 2.3 × 0.2641720524 gal

=0.60759572 gal

Hence , 0.60759572 gal is the volume of acid in gallons  in a sample of hydraulic acid .

Learn more about gallon ,here:

brainly.com/question/14296555

#SPJ1

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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
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Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

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Answer:

1.71x10²⁷

Explanation:

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C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

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+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

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Why does a red object appear red?
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