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Stella [2.4K]
3 years ago
15

What did Zoya and her freind do in response

Chemistry
1 answer:
Mama L [17]3 years ago
5 0
In response of what like what’s the full clear question
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What is the total number of moles represented by 20 grams of CACO3
Nutka1998 [239]

Answer:

B. 0.2.

Explanation:

  • We can use the relation:

<em>n = mass/molar mass</em>

mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.

<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>

<em></em>

<em>So, the right choice is: B. 0.2.</em>

8 0
3 years ago
What can occur in physical change ?
Alexandra [31]

Within a physical change, an element can change forms, such as going from solid to a liquid through melting. Color change can also occur during a physical change. Physical changes are very different from chemical changes. In a chemical change the element itself changes into something else within a reaction, such as combustion (burning).


Hope this helped

5 0
3 years ago
What is the relationship between atoms and static electricity
Mnenie [13.5K]
They both build up to form electricity
3 0
3 years ago
Read 2 more answers
Which criteria determine whether a heterogeneous mixture is a colloid or a suspension
Artemon [7]
<span>whether the particles do not settle for an extended period of time</span>
6 0
3 years ago
The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca
geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

M=d\times V=1.00 g/mL\times 150 ml =150.0 g

Mass of solution = m = M = 150.0 g

NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol

Moles of NaOH = \frac{5.00 g}{40 g/mol}=0.125 mol

Energy released when 0.125 moles of NaOH added in water = Q

Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

Q'= 5,563.8 J

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = \Delta T

Q'=mc\times \Delta T

5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T

\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0
3 years ago
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