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Aleks [24]
3 years ago
6

When aqueous solutions of K2CO3 and Cu(NO3)2 are mixed, what precipitate(s) will form?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
7 0
When aqueous solutions of Potassium Carbonate and Copper Nitrate are mixed, the potassium will displace the copper, as it is more reactive, and Potassium Nitrate and Copper Carbonate will be formed.

The equation for this reaction is (ignoring state symbols):

K₂CO₃ + Cu(NO₃)₂ ⇒ 2KNO₃ + CuCO₃
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The transfer of energy by waves is also known as _______
Verdich [7]

Answer:

wave

Explanation:

'Wave' is a common term for a number of different ways in which energy is transferred: In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields.

i hope this helps and your welcom

5 0
3 years ago
A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos
Rama09 [41]

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

3 0
3 years ago
What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific he
atroni [7]
LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.
4 0
3 years ago
Calculate the standard enthalpy of formation of carbon disulfide (CS2) from it's elements, given that C(graphite) + O2(g) → CO2(
ch4aika [34]

Answer:

Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol

Explanation:

forming CS2 means that it should in the product side

C(graphite) + O2 → CO2                  ΔH = -393.5

2S(rhombic) + 2O2 → 2SO2            ΔH = -296.4 x 2

CO2 + 2SO2 → CS2 + 3O2             ΔH = -1073.6 x -1

the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.

the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.

after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2

Add ΔH to find the enthalpy of formation of CS2

ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol

be aware of signs

3 0
3 years ago
What is one advantage of using a computer over a graphing calculator
CaHeK987 [17]

PCs are usually more powerful devices w/ higher resolutions than the common graphing calculator.

3 0
4 years ago
Read 2 more answers
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