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Reptile [31]
3 years ago
6

subtract the mass of the filter paper (0.27 g) from the mass of the paper and copper (0.98 g). Record the difference in the data

table as the amount of copper.
Chemistry
1 answer:
vova2212 [387]3 years ago
7 0
When measuring copper that is collected on a filter paper, the mass measurement obtained will include both the mass of the filter paper and copper.
therefore we have to find the mass of the empty filter paper before copper is on the paper.
the mass of the filter paper alone is 0.27 g
then the mass of copper and filter paper is 0.98 g
therefore to find the mass of copper alone we have to take the difference between the 2 masses
mass of copper = (mass of copper + filter paper ) - mass of filter paper
                          = 0.98 g - 0.27 g 
mass of copper = 0.71 g
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A 5.00 gram sample of magnesium sulfate hydrate (epsom salt) is heated in the lab to form anhydrous magnesium sulfate. After hea
Ostrovityanka [42]

Answer:

The answer to your question is  MgSO₄ 5H₂O

Explanation:

Data

mass of MgSO₄ = 2.86 g

mass of H₂O = 2.14 g (5 - 2.86)

Process

1.- Calculate the molecular mass of the compounds

MgSO₄ = 24 + 32 + (16 x 4) = 120

H₂O = 16 + 2 = 18

2.- Convert the grams obtain to moles

                        120 g of MgSO₄  --------------- 1 mol

                         2.8 g                  ----------------  x

                         x = (2.8 x 1)/120

                        x = 0.024 moles

                         18 g of H₂O --------------------- 1 mol

                         2.14 g           -------------------- x

                         x = (2.14 x 1)/18

                         x = 0.119

3.- Divide by the lowest number of moles

MgSO₄  = 0.024/0.024 = 1

H₂O = 0.119/ 0.024 = 5

4.- Write the molecular formula

                   MgSO₄5H₂O                        

5 0
4 years ago
What is the enthalpy for reaction 1 reversed? reaction 1 reversed: co2→co + 12o2 express your answer numerically in kilojoules p
QveST [7]
The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:

CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol

As observed O2 will not have enthalpy of formation as it is a pure substance. 

To calculate for the enthalpy of reaction,
   enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
    = 294.5 kJ/mol

ANSWER: 294.5 kJ/mol
8 0
4 years ago
A gas has a volume of 5.0 L at a pressure of 50 KPa. What happens to the volume when the pressure is increased to 125?
Alexeev081 [22]
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2 
P is pressure and V is volume.
P1 = 50     P2 = 125
V1 = 5       V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.

7 0
3 years ago
Can sombody pls help me with this​
luda_lava [24]
Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)
6 0
3 years ago
Why is the solvation energy negative?<br><br> a) bonds are being created<br> b) bonds are breaking
IrinaVladis [17]

Answer:

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3 0
3 years ago
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