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igor_vitrenko [27]
3 years ago
15

HELP!

Chemistry
1 answer:
vova2212 [387]3 years ago
5 0

Answer:

14.53

Explanation:

V1= 218ML

V2= ?

P2= 15P1

USING BOYLE'S LAW

P1V1= P2V2

V2= P1V1/P2

V2 = P1(218ML)/15P1 = 14.53ML

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The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

6 0
3 years ago
What mass of H2SO, is required to react completely with 50.0 grams of ammonia (NH3)?​
Reika [66]
The answer is 10.7 grams
8 0
3 years ago
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
borishaifa [10]

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

3 0
3 years ago
How many atoms doesCo(NO3)2 have
svet-max [94.6K]
The answer is 9. Calcium nitrate is made up
of 3 different elements that equal 9 atoms.
7 0
3 years ago
Read 2 more answers
In the laboratory, a quantity of I2 was reacted with excess H2 to give 1.26 moles of HI. It is also known that the percent yield
Anon25 [30]

Answer:

1.008moles of iodine

Explanation:

Hello,

This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.

Percent yield = (actual yield / estimated yield) × 100

Actual yield = 1.2moles

Estimated yield = ?

Percentage yield = 84%

84 / 100 = 1.2 / x

Cross multiply and solve for x

100x = 84 × 1.2

100x = 100.8

x = 100.8/100

x = 1.008moles

1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI

5 0
4 years ago
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