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vaieri [72.5K]
3 years ago
15

Under each photo,

Chemistry
1 answer:
cestrela7 [59]3 years ago
8 0

Answer:

Explanation:

water vapor:in a cloud or when you breath out once it's cold.Ice: in your freezer or on a frozen lake.liquid water:at a stream

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Calculate the concentration 3.8g of copper sulfate, CuSO4 dissolved in 250cm4 of water
11Alexandr11 [23.1K]

Answer:

Molarity = 0.08 M

Explanation:

Given data:

Mass of copper sulfate = 3.8 g

Volume of water = 250 cm³  (250/1000 = 0.25 L)

Concentration of solution = ?

Solution:

Number of moles of copper sulfate:

Number of moles = mass/molar mass

Number of moles = 3.8 g/ 159.6 g/mol

Number of moles = 0.02 mol

Concentration:

Molarity = Number of moles / volume in L

By putting values,

Molarity = 0.02 mol / 0.25 L

Molarity = 0.08 mol/L

Molarity = 0.08 M

6 0
3 years ago
For the reaction system, h2(g) + x2(g) ↔ 2 hx(g), kc = 24.4 at 300 k. a system made up from these components which is at equilib
Pavel [41]
Kc=24.4=[HX]∧2/[H2]×[X2] =(0.6)∧2/(0.2)×[H2]
[H2] = 0.36/(24.4×0.2) = 0.07377 mole
4 0
4 years ago
Read 2 more answers
A. Look at the inheritance patterns. What do you notice?
mario62 [17]

Answer:

Explanation:

i notice that theri is no image so i wont be able to help you sorry

5 0
3 years ago
What volume (ml) of 3.0 m naoh is required to react with 0.8024-g copper(ii) nitrate? what mass of copper(ii) hydroxide will for
babunello [35]
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂

Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g 
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
 - 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2 
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction

Q2) 
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up. 
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of  1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol 
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g 
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
7 0
4 years ago
Read 2 more answers
4. Complete the following equations:
lara31 [8.8K]

Answer:

2NaCl + CuCO3

FeCl2 + BaSO4

CuCO3 + Ca(NO3)2

Explanation:

<em>Presumably this is a double replacement reaction.</em>

<em>A+B</em><em> </em><em>+</em><em> </em><em>C</em><em>+</em><em>D</em><em> </em><em>→</em><em> </em><em>A+D</em><em> </em><em>+</em><em> </em><em>C+B</em>

<em>It</em><em> </em><em>seems</em><em> </em><em>I</em><em> </em><em>may</em><em> </em><em>be</em><em> </em><em>wrong</em><em> </em><em>so</em><em> </em><em>please</em><em> </em><em>try</em><em> </em><em>to</em><em> </em><em>work</em><em> </em><em>out</em><em> </em><em>the</em><em> </em><em>problem</em><em> </em><em>yourself</em><em> </em><em>to</em><em> </em><em>double</em><em> </em><em>check</em><em>,</em><em> </em><em>keeping</em><em> </em><em>in</em><em> </em><em>mind</em><em> </em><em>the</em><em> </em><em>charges</em><em> </em><em>of</em><em> </em><em>each</em><em> </em><em>compound</em><em>.</em>

4 0
3 years ago
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