All categories

3 years ago

Water a combination of the elements hydrogen and oxygen, is an example of an

Chemistry

1 answer:

marin [14]3 years ago

3 0

Answer:

Explanation:

Common Examples of the Law of Definite Proportions

Water, written as the chemical compound H20, is made up of atoms of hydrogen and oxygen. If one oxygen atom is combined with two hydrogen atoms, water is created.

You might be interested in

Is the igneous rock destructive or constructive?

sammy [17]

This certain type of rock is constructive, not only from the fact that it was made up from earthly habits. But also it breaks down very easily, which can help the environment. Therefore a sedimentary rock is constructive.

5 0

3 years ago

How many grams of iron are needed to produce 75.0 g of iron(III) chloride?

OleMash [197]

Answer:

Explanation:

Hello!

In this case, given the chemical reaction:

In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:

It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:

8 0

3 years ago

What’s the difference between a physical and chemical property

irga5000 [103]

Answer:

One is hands on and one is watching

Explanation:

I just know my science

3 0

3 years ago

Which statement correctly describes the location and charge of the protons in an atom?

Bezzdna [24]

Answer:

Protons are located in the nucleus and have a positive charge and a mass of 1.

Explanation:

Know about protons, neutrons, and electrons.

6 0

3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}