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3 years ago

How many liters of 15 acid and 33 acid should be mixed to make 40 liters of 21 acid solution answer key?

1 answer:

6 0

Answer is: 26.67 liters of 15% acid and 13.33 liters of 33% acid are needed.
ω₁(acid) = 15% = 0,15.
ω₂(acid) = 33% = 0,33.
V₁(acid) = ?
V₂(acid) = ?
V₃(acid) = 40l.
ω₃(acid) = 21% = 0,21
1) V₁ + V₂ = 40l, V₁ = 40 - V₂.
2) 0,15·V₁ + 0,33·V₂ ÷ 40 = 0,21
0,15·(40 - V₂) + 0,33·V₂ ÷ 40 = 0,21
6 - 0,15V₂ + 0,33V₂ ÷ 40 = 0,21.
0,18V₂ = 2,4, V₂ = 13,33l.
V₁ = 40 - 13,33 = 26,67l.

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3 years ago

Which metal was more reactive in water? How do you know?

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3 years ago

Read 2 more answers

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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3 years ago

A 1.59 mol sample of Kr has a volume of 641 mL. How many moles of Kr are in a 4.41 L sample at the same temperature and pressure

Marina86 [1]

Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

n₁= 1.59 mol

V₁= 641 mL= 0.641 L

For the final gas (2), we have:

V₂: 4.41 L

The relation between 1 and 2 is given by:

n₁/V₁ = n₂/V₂

We calculate n₂ as follows:

n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol

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2 years ago