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3 years ago

PLEASE HELP!!! 10 POINTS

Chemistry

1 answer:

k0ka [10]3 years ago

3 0

i think the answer is C

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The nh2cl model consists of balls and sticks. the electrons in the molecule are paired, and each stick represents a valence elec

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4 years ago

Production of Jam from Crushed Fruit in Two Stages. In a process producing jam (C1), crushed fruit containing 14 wt % soluble so

Andrej [43]

Answer:

The total kg from the mixer: $m=2222.5 kg$

Evaporated water: $m_{water ev.}=791.1 kg$

Jam produced: $m_{jam}=1431.4 kg$

Explanation:

Step by step:

1) Kg of mixture from the mixer

It says that crushed fruit is added to the mixer with the sugar and pectin.

The fruit added is 1000 kg

The sugar added is 1.22 kg sugar/1 kg crushed fruit:

$m_{sugar}=1000 kg fruit * \frac{1.22 kg sugar}{1 kg fruit}=1220 kg sugar$

The pectin added is 0.0025 kg pectin/1 kg crushed fruit:

$m_{pectin}=1000 kg fruit * \frac{0.0025 kg pectin}{1 kg fruit}=2.5 kg pectin$

The total kg from the mixer:

$m=1000 kg fruit + 2.5 kg pectin + 1220 kg sugar=2222.5 kg$

2) Evaporated water

To calculate the evaporated water it's important to have in mind that the mix goes from a concentration of 14 wt% to 67 wt%. This difference is because of the evaporated water. So:

Initial: 1000 kg of fruit with 14 wt% of solids

$m_{solids}=1000 kg *0.14=140 kg$

<em>This amount of solids is constant </em>

Final: The mass of solids is the same but now it represents the 67 wt%

$m_{final}=140 kg *\frac{100}{67}=208.9 kg$

and

$m_{water ev.}=1000kg-m_{final}$

$m_{water ev.}=1000kg-208.9kg=791.1 kg$

3) Jam produced

$m_{jam}=m_{final}+m_{pectin}+m_{sugar}=208.9 kg + 1220 kg + 2.5 kg$

$m_{jam}=1431.4 kg$

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3 years ago