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Ludmilka [50]
3 years ago
15

Which values are written in proper scientific notation? Check all that apply. –350 –350.00 –3.50 × 104 0.000350 3.50 × 104 3.50

× 10–4 35.0 × 10–5 3,500
Chemistry
2 answers:
Oksanka [162]3 years ago
7 0
-3.50 * 10^4
3.50 * 10^4
3.50 * 10^-4
algol133 years ago
7 0

Answer : The correct options are, -3.50\times 10^{4},3.50\times 10^{4}\text{ and }3.50\times 10^{-4}

Explanation :

Scientific notation : The way or representation of expressing the number that are too big or too small that is written in the decimal form. That means always written in the power of 10 form.

For example : 700 is written in scientific notation as, 7.0\times 10^2

According to the question, -3.50\times 10^{4},3.50\times 10^{4}\text{ and }3.50\times 10^{-4}  are written in proper scientific notation. While the other options are not written in proper scientific notation.

Therefore, the correct options are, -3.50\times 10^{4},3.50\times 10^{4}\text{ and }3.50\times 10^{-4}

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A 0.1025-g sample of copper metal is dissolved in 35 ml of concentrated hno3 to form cu2 ions and then water is added to make a
GalinKa [24]

Molarity is defined as number of moles of solute in 1 L of solution.

Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.

The balanced reaction will be as follows:

Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O

From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:

n=\frac{m}{M}

molar mass of Cu is 63.55 g/mol thus,

n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol

Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:

M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M

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7 0
3 years ago
A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Zn2+ solution
Lana71 [14]

Explanation:

The given reaction at cathode will be as follows.

At cathode: Zn^{2+} + 2e^{-} \rightarrow Zn,     E_{o} = -0.761 V

At anode: Zn \rightarrow Zn^{2+} + 2e^{-},       E_{o} = 0.761

Therefore, net reaction equation will be as follows.

                 Zn^{2+} + Zn \rightarrow Zn + Zn^{2+}

Initial:     0.129         -            -       0.427

Change:  -0.047      -            -     -0.047

Equilibrium: (0.129 - 0.047)      (0.427 - 0.047)

                = 0.082                       = 0.38

As E^{o}_{cell} for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

                   E_{cell} = E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}

                              = 0 - \frac{8.314 atm L/mol K \times 291 K}{2 \times 96500} ln \frac{0.38}{0.082}

                              = 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.

3 0
4 years ago
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