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Lina20 [59]
3 years ago
5

How many valence electrons are in the element copernicium? Will GIVE BRainLYEsT!!!!!

Chemistry
2 answers:
kirill [66]3 years ago
7 0
7s ^2 it will have 2 valence electrons
Ivahew [28]3 years ago
3 0
As it is located at 7s ^2 it will have 2 valence electrons that due to its position in the s orbital it will be prone to losing them to obtain a noble gas configuration.
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Use the diagram to determine: How many atoms are on the Right side =___ How many atoms are on the Left side = ____ Is this a Bal
Lady_Fox [76]

Answer:

on the right side two and on left side it is6 yes it is a balanced equation pls mark me as the brainliset hope it helps you

5 0
3 years ago
The sun causes a greenhouse effect on Earth. How is Earth impacted by the greenhouse effect?(1 point)
creativ13 [48]

Answer: Sunlight passes through the atmosphere and warms the Earth's surface. ... As more greenhouse gases are emitted into the atmosphere, heat that would normally be radiated into space is trapped within the Earth's atmosphere, causing the Earth's temperature to increase.

Explanation: This keeps well life on earth so the answer is (3)

:)

4 0
2 years ago
How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?
pav-90 [236]
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
8 0
3 years ago
Does anyone know the answer to these?
Rashid [163]

Answer:

The answer to your question is below

Explanation:

Data

mass of CaCO₃ = 155 g

mass of HCl = 250 g

mass of CaCl₂ = 142 g

reactants = CaCO₃ + HCl

products = CaCl₂ + CO₂ + H₂O

1.- Balanced chemical reaction

             CaCO₃ + 2HCl   ⇒    CaCl₂ + CO₂ + H₂O

2.- Limiting reactant

molar mass of CaCO₃ = 40 + 12 + 48 = 100 g

molar mass of HCl = 2[1 + 35.5 ] = 73 g

theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37

experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃

3.-

Calculate the molar mass of CaCl₂

CaCl₂ = 40 + 71 = 111 g

          100 g of CaCO₃ ------------------ 111 g of CaCl₂

           155 g of CaCO₃ ----------------- x

               x = (155 x 111) / 100

               x = 17205 / 100

              x = 172.05 g of CaCl₂

4.- percent yield

Percent yield = 142 / 172.05 x 100 = 82.5 %

5.- Excess reactant

    100 g of CaCO₃  -------------------- 73 g of HCl

     155 g of caCO₃ ------------------- x

           x = (155 x 73)/100

           x = 133.15 g

Mass of HCl = 250 - 133.15

                    = 136.9 g

4 0
3 years ago
Question:<br> What is the molar concentration of 1.29 mol of KCL dissolved in 350 mL of solution?
STatiana [176]

Answer:

M = 3.69 M.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the molar concentration of the 1.29 moles of KCl in 350 mL of solution by recalling the mathematical definition of molarity as the division of the moles by the volume in liters, in this case 0.350 L; thus, we proceed as follows:

M=\frac{1.29mol}{0.350L}\\\\M=3.69M

Which gives molar units, M, or just mol/L.

Regards!

8 0
3 years ago
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