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3 years ago

11

In HNO3, the oxidation state of hydrogen is +1 and the oxidation state of oxygen is -2. Therefore, the oxidation state of nitrog

en is -

1 answer:

marta [7]3 years ago

4 0

+1 Because of all the numbers of charges -2+1+x=0 so, the number should be 1 for the oxidation of H.

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Appropriate units for the speed of a chemical reaction, the reaction rate, are _______.

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Appropriate units for the speed of a chemical reaction, the reaction rate, are M/s

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3 years ago

What happens if the metal you throw in is MORE REACTIVE than the<br> metal ion in solution?

IRISSAK [1]

A displacement reaction Would occur in this situation

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

Help me ASAP PLS

n200080 [17]

Answer:

18 g

Explanation:

We'll begin by converting 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 0.2 M

Mole of C₆H₁₂O₆ =?

Molarity = mole / Volume

0.2 = Mole of C₆H₁₂O₆ / 0.5

Cross multiply

Mole of C₆H₁₂O₆ = 0.2 × 0.5

Mole of C₆H₁₂O₆ = 0.1 mole

Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.1 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mass = mole × molar mass

Mass of C₆H₁₂O₆ = 0.1 × 180

Mass of C₆H₁₂O₆ = 18 g

Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.

6 0

3 years ago

Help me please show the work

Feliz [49]

2 => 6 meters = 600cm

3 => 70 quarts = 17.5 gallon

50 inches = 1.5 meters

2,000 grains = 4.57 ounces

7 tablespoons = 0.027344 gal

3 0

3 years ago