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Alja [10]
3 years ago
14

Write the reaction rate expressions for the reaction below in terms of the disappearance of the reactants and the appearance of

products. Give the expressions for the disappearance of the reactants first, in the order written in the chemical equation. Then write the expressions for the appearance of the products in the order written in the chemical equation. Write the expressions in order of appearance in the equation in the form. ± 1 X × Δ[Α] Δt where ± is either a plus OR a minus sign, not both, X is an integer, and A is a chemical species. Do not include the state of matter. 2H2(g) + O2(g) → 2H2O(g)
Chemistry
1 answer:
wlad13 [49]3 years ago
6 0

Answer: rate of disappearance of hydrogen = -\frac{1d[H_2]}{2dt}   rate of disappearance of oxygen =  -\frac{1d[O_2]}{1dt}

rate of appearance of water = +\frac{1d[H_2O]}{2dt}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2H_2(g)+O_2(g)\rightarrow 2H_2O(g)

The rate law is:

Rate=k[H_2]^2[O_2]^1

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

rate of disappearance of hydrogen=  -\frac{1d[H_2]}{2dt}

rate of disappearance of oxygen = -\frac{1d[O_2]}{1dt}

rate of appearance of water = +\frac{1d[H_2O]}{2dt}

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At what temperature would a 1.30 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86
Setler [38]

Answer:

-4.59°C

Explanation:

Let's see the formula for freezing point depression.

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Freezing constant. For water if 1.86°C/m

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor.

0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9

Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)

Freezing T° of solution = - (1.86°C/m .  1.30m . 1.9) → -4.59°C

NaCl →  Na⁺  +  Cl⁻

4 0
3 years ago
A kilogram is an SI unit. Which quantity might you measure in kilograms? temperature
jonny [76]
The quantity that you would measure in kg, is mass.
4 0
3 years ago
Read 2 more answers
How many joules of heat are needed to raise the temperature of 47.5 g of aluminum from 21°C to 94°C , if the specific heat of al
Montano1993 [528]

Answer:

3120.75J

Explanation:

So, we have the formula q = mc\Delta t. For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in \frac{J}{{g\°C}}, and \Deltat is the change in temperature. In this case, m = 47.5g, c = 0.9  \frac{J}{{g\°C}}, and \Deltat  = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be \frac{J}{{g\°C}}. I may be in Kelvin or something. Anyways, hope this helps.

5 0
4 years ago
A geochemist measures the concentration of salt dissolved in Lake Parsons and finds a concentration of 21.0 gL⁻¹. The geochemist
dedylja [7]

Answer:

The percentage of Lake Parsons which has evaporated since it became isolated is 68.24%.

Explanation:

The concentration of salt dissolved in Lake Parson = 21.0 g/L

= In 1 L of lake Parson  water = 21.0 g of salt

The concentration of salt in several nearby non-isolated lakes = 6.67  g/L

The salt concentration in Lake Parsons before it became isolated = 6.67 g/L

In 1 L of lake water = 6.67 g of salt

If 1 L of lake Parson  water has 6.67 g of salt. Then 21.0 grams of salt will be in ;

\frac{1}{6.67}\times 21.0 L = 3.15 L

Water evaporated = 3.15 L - 1 L = 2.15 L

The percentage of Lake Parsons which has evaporated since it became isolated:

\frac{2.15 L}{3.15 L}\times 100=68.25\%

5 0
4 years ago
The formula B+ O2_ B2O3 is balanced as
julsineya [31]

Hey there!

B + O₂ → B₂O₃

Balance O.

2 on the left, 3 on the right. Add a coefficient of 3 in front of O₂ and a coefficient of 2 in front of B₂O₃.

B + 3O₂ → 2B₂O₃

Balance B.

1 on the left, 4 on the right. Add a coefficient of 4 in front of B.

4B + 3O₂ → 2B₂O₃

Our final balanced equation: 4B + 3O₂ → 2B₂O₃

Hope this helps!

5 0
4 years ago
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