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Kamila [148]
3 years ago
5

How to measure the rate of a chemical reaction​

Chemistry
1 answer:
tiny-mole [99]3 years ago
3 0
Reaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt. The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.
If a reaction produces a gas such as oxygen or carbon dioxide, there are two ways to measure the reaction rate: using a gas syringe to measure the gas produced, or calculating the reduction in the mass of the reaction solution.
If the reaction produces a precipitate, the amount formed can be used to determine reaction rate by measuring how long it takes for the forming precipitate to obscure the visibility of a cross through a conical flask.
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Page. 5. What is the position of alkali metals in the Modern periodec table. ​
hoa [83]

Answer:

Group 1 ( The Leftmost)

More Information :

The alkali metals are six chemical elements in Group 1, the leftmost column in the periodic table.

They are :

  • lithium (Li)
  • sodium (Na)
  • potassium (K)
  • rubidium (Rb)
  • cesium (Cs)
  • francium (Fr).
6 0
3 years ago
Read 2 more answers
1. How much heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C?
diamong [38]

Answer: An amount of 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

Explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

1 kg = 1000 g\\4.64 kg = 4.64 kg \times \frac{1000 g}{1 kg}\\= 4640 g

T_{1} = 150^{o}C

T_{2} = 219^{o}C

The standard value of specific heat of lead is 0.128 J/g^{o}C.

Formula used to calculate heat is as follows.

q = m \times C \times \Delta T

where,

q = heat energy

m = mass of substance

C = specific heat of substance

\Delta T = change in temperature

Substitute the value into above formula as follows.

q = m \times C \times \Delta T\\= 4640 g \times 0.128 J/g^{o}C \times (219 - 150)^{o}C\\= 40980.48 J/g^{o}C

Thus, we can conclude that 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

8 0
3 years ago
The solubility of acetanilide in hot water (5.5 g/100 ml at 100∘C) is not very great, and its solubility in cold water (0.53 g/
Olin [163]

Answer:

89.4%

Explanation:

Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.

So, the mass that is recovered is 5.0 - 0.53 = 4.47 g

The percent recovery is:

(4.47/5)x100% = 89.4%

8 0
4 years ago
You are comparing a reaction that produces a chemical
MaRussiya [10]

If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.

7 0
4 years ago
Read 2 more answers
H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a
forsale [732]

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = \frac{[SO42-] [H3O+]}{[HSO4-]}

0.011 = \frac{x (0.1 + x)}{0.1 - x}

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = { -b +,-\sqrt{b^2 - 4ac / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = \sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }  / 2(-1)

x = 0.111 +,- \sqrt{0.012321 + 0.0044} / -2

x = 0.111 +,- \sqrt{0.016721} / -2

x = \frac{0.111 +, - 0.1293}{-2}

x = \frac{0.111 + 0.1293}{-2}   , x = \frac{0.111  - 0.1293}{-2}

x = \frac{0.2403}{-2}    , x = \frac{0.0183}{-2}

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

8 0
3 years ago
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