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gladu [14]
3 years ago
10

A decrease in temperature will affect the rate of which of the following reaction

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
6 0
You didn't put the reactions
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What are two ways that scientists u measure temperature
anygoal [31]

Using the same anchor points, he determined the freezing temperature for water to be 0 degree and the boiling temperature 100 degrees. The Celsius scale is known as a Universal System Unit. It is used throughout science and in most countries.


8 0
3 years ago
a 9.84 ounces ingot of unknown metal is heated from 73.2 degrees fahrenheit - 191.2 degrees fahrenheit this requires 3.912 calor
Gekata [30.6K]

The SI unit of specific heat is J per gram per degree Celsius. Thus it follows that specific heat could be calculated in this way:

Specific Heat = Energy / (mass x change in temperature)

Thus,

Specific Heat = 3.912 cal / (9.84 oz x (191.2 ˚F – 73.2 ˚F))

Specific Heat = 3.369 x 10^-3 cal/oz-˚F

6 0
4 years ago
Which of the following models best describes the arrangement of valence electrons in metals?
Lana71 [14]

  The arrangement of valence electrons in metals can be best described by the

A sea of electrons.

7 0
3 years ago
The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear
djverab [1.8K]

Answer: The rate of appearance of Br_2 is 0.0700Ms^{-1}

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.

2HBr(g)\rightarrow H_2(g)+Br(g)

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 in terms of disappearance of HBr = -\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2 = \frac{1d[H_2]}{dt}

Rate in terms of appearance of Br_2 = \frac{1d[Br_2]}{dt}

-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}

Given :

-\frac{1d[HBr]}{dt}=0.140Ms^{-1}

The rate of appearance of Br_2;

\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}

Thus rate of appearance of Br_2 is 0.0700Ms^{-1}

6 0
3 years ago
Each elements on the periodic chart is made up of a certain kind of atom. Agree or Disagree reason why
shutvik [7]
All atoms are made up of different amounts of atoms. Ex: some may have two some may have 3
4 0
4 years ago
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