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geniusboy [140]
4 years ago
6

Consider three devices operating connected at steady state: a pump, a boiler, and

Chemistry
1 answer:
MA_775_DIABLO [31]4 years ago
3 0
C) the heat transfer to the boiler
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Igoryamba
B or D but id say go with D

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3 years ago
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What is the difference between covalent bonding and electrovalent bonding​
Rudiy27

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1 ) Electrovalent compounds are formed by the complete transfer of electrons while covalent compounds are formed by sharing of electrons between 2 atoms.

2) Electrovalent compounds are more soluble in polar solvents like water while covalent compounds are more soluble in non-polar solvents like methane.

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5 0
3 years ago
When 25.0 ml of 0.500 m h2so4 is added to 25.0 ml of 1.00 m koh in a coffee-cup calorimeter at 23.50°c, the temperature rises to
AlladinOne [14]

The chemical reaction that occurs between the given substances is a neutralization reaction as shown below :

 

H_2SO_4 + 2KOH -> K_2SO_4 + 2H_2O

<span>1 mol            2 mol      1 mol           2 mol </span>

 

<span>The number of moles of the given substances  is calculated as shown : </span>

Number of moles of H_2SO_4 = 25.0 mL x 0.50 M = 12.5 millimoles

Number of moles of KOH = 25.0 mL x 1.00 M = 25.0 millimoles

 

As 1 mol of sulfuric acid reacts with 2 mol of KOH to give 2 mol of water, 12.5 millimoles of sulfuric acid completely reacts with 25.0 millimoles of KOH to give 25.0 millimoles of water.

Total volume of the solution = 25.0 mL + 25.0 mL = 50.0 mL.

Density of water is 1 g/mL. Use this to calculate the mass of the solution.

<span>Mass of the solution – 50.0 mL x 1 g/mL = 50.0  </span>

The specific heat of water is 4.184 J/gK. The temperature of the solution is increased from 23.5 degrees Celsius to 30.17 degrees Celsius.

The amount of heat released = 4.184 J/gK x 50.0 g x (30.17C – 23.50C) 1395 J

 

The amount of heat released per one mole of water formed can be calculated as shown :

The amount of heat released for formation of mole water = 1395 J / (25.0 m mol x 1mol/1000 m mol)

= 55,800 J

<span> </span>

8 0
3 years ago
1. Calculate the quantity of glucose needed to prepare (a) 250 mL of a 100 mM of
weeeeeb [17]

Answer:

a. mass of glucose required = 4.50 g

b. mass of glucose required = 0.05 g

c. mass of glucose required = 25g

Explanation:

1a. molar mass of glucose = 180.156 g/mol

volume of solution = 250 mL or 250 mL * 1 L/1000 mL = 0.25 L

concentration of solution = 100 mM or 100mM * 1 M/1000 mM = 0.10 M

amount in moles = mass/ molar mass = concentration * volume

mass in grams = concentration * volume * molar mass

mass = 0.10 * 0.25 * 180.156

mass of glucose required = 4.50 g

b. mass concentration of 0.05 % solution = 0.05 g/100 ml = 0.5 g/L

volume of solution = 100 mL = 0.1 L

mass in grams = mass concentration * volume

mass in grams =  0.5 g/L * 0.1 L

mass of glucose required = 0.05 g

c. mass concentration of solution = 50 mg/ mL = 0.05 g/ 0.001 L = 50 g/L

volume of solution = 500 mL = 0.5 L

mass of solute required = concentration * volume

mass required = 50 g/L * 0.5 L

mass of glucose required = 25g

7 0
3 years ago
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