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zlopas [31]
3 years ago
15

Please help go over my work

Chemistry
2 answers:
maxonik [38]3 years ago
4 0

Answer:

it's A

Explanation:

......................

beks73 [17]3 years ago
3 0
It’s a idjdjjdjdjjfkdkdkdjjdfjjdhshdh
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Write the balanced half equation of iron 2 and permanganate in a solution of acid. Show all of your work.
Aleks [24]

Answer:

5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

Explanation:

Fe⁺² + MnO₄⁻ +  H⁺ => Mn⁺² + Fe⁺³ +  H₂O

5(Fe⁺² => Fe⁺³ + 1e⁻)      =>                        5Fe⁺² => 5Fe⁺³ + 5e⁻

<u>MnO₄⁻ + 5e⁻ => Mn⁺²    =>   MnO₄⁻ + 8H⁺ + 5e⁻ => Mn⁺² + 4H₂O</u>

                                      =>  5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

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3 years ago
Which reflects more sunlight-dark soil or white sand?
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Which element(s) is/are not balanced in this equation?
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What are the ingredients in Swedish fish
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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
3 0
3 years ago
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