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TEA [102]
3 years ago
10

What is the mass, volume, density, and substance for block D:

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

EXPLANATION :

mass is the quantity of matter in an object .

volume is the amount of space that something contains or fills .

density is the thickness of a solid , liquid or gas measured by its mass per unit of volume.

Thanks I did what I can hope it helped...

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What is the mass of one mole of Ca(OH)2?
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Answer:

74.10g/mol

Explanation:

The mass of one mole of Ca(OH)₂ is the molar mass of the compound.

To solve this problem, we add the atomic masses of the species in the compound;

 Atomic mass of Ca = 40.078g/mol

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So;

  Molar mass  =  40.078 + 2(15.9994 +1.0079 )  = 74.10g/mol

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3 years ago
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How many of the organisms represented are multicellular
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5 0
3 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

6 0
3 years ago
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