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2 years ago

The rate of disappearance of HCl was measured for the following reaction:

Chemistry

1 answer:

AURORKA [14]2 years ago

6 0

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

after 430mins,t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

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Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given

professor190 [17]

Answer: -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.

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2 years ago

Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m na

Fed [463]

The molarity of acetic acid in the vinegar is 0.94 M

calculation

Step 1: write the balanced equation between CH3COOH + NaOH

that is CH3COOH + NaOH → CH3COONa + H2O

step 2 : find the moles of NaOH

moles =molarity x volume in L

volume in liters = 26.90/1000=0.0269 l

moles = 0.175 mol /L x 0.0269 L =0.0047 moles of NaOH

Step 3: use the mole ratio to find moles of CH3COOH

that is the mole ratio of CH3COOH: NaOH is 1:1 therefore the moles of CH3COOH is =0.0047 moles

Step 4: find the molarity of CH3COOH

molarity = moles/volume in liters

volume in liter = 5.00/1000 =0.005 l

molarity is therefore=0.0047 moles/ 0.005 l = 0.94 M

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2 years ago

Analyze the example of this door knob wheel and axle.

bulgar [2K]

Answer:

Explanation:

Velocity ratio= Radius of wheel/radius of axle

Radius of wheel= 4.125 inches

Radius of axle= 0.125 inches

Velocity ratio = 4.125/0.125 = 33

Then;

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Since the efficiency of the system = 85%

85 = mechanical advantage/33 × 100

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2 years ago

Olive oil has a density of 0.92g/ml. How much would 1.0 Liter of olive oil weigh in grams?

Julli [10]

Answer:

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Explanation:

Data obtained from the question include the following:

Density = 0.92g/ml

Volume = 1L = 1 x 1000 = 1000mL

Mass =..?

Density is simply defined as the mass of the substance per unit volume of the substance. Mathematically it can be represented as:

Density = Mass /volume.

Mass = Density x volume

Mass = 0.92 x 1000

Mass = 9.2x10²g.

Therefore, 1L of olive will weigh 9.2x10²g.

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galina1969 [7]

Answer:

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