Calculate the wavelength in meters of electromagnetic radiation that has a frequency of 1.09 × 10⁸ s⁻¹. (c = 3.00 X 10⁸ m/s)

Let's say we have 2 Zinc atoms and 2 Sulfur atoms on one side of the equation. According to the Law of Conversation and Mass, ho

Viefleur [7K]

Answer:

Two

Explanation:

The law of conservation of mass states that in an isolated system the mass present is neither destroyed nor created by chemical changes or physical changes.

This tells us that the mass of reactants must be equal to the product mass.

If 1 atom of Zn react with one atoms of sulfur, the product will be 1 molecule of zinc sulfide according to the equation below

Zn(s) + S(s) ⇒ ZnS(s)

therefore two atoms each f zinc and sulfur will product two molecules of Zn sulfide

5 0

2 years ago

A 0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191Â g of co

olchik [2.2K]

The chemical formula for the compound can be written as,

CxHyOz

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,

CxHyOz + O2 --> CO2 + H2O

number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H

Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
C3/2H3/2O

The empirical formula of the compound is therefore,
 C₃H₃O₂

5 0

2 years ago

Salim takes 100 g of water each in four identical containers P, Q, R and S and mixes the following in them. 5 g of salt in conta

Simora [160]

The container with chalk powder will contain the least amount of water, because it absorbs water, but the containers with honey and cocunut oil will conserve their amount of water, because they will prevent the water evaporation (especially cocunut oil because it will be on the top side of the container).

7 0

2 years ago

Determine the number of moles and mass requested for each reaction in Exercise 4.42.

suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

Step 1: Write the balanced equation
Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$ [/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$

5 0

2 years ago

Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A

lesantik [10]

Answer: The chemical equations and equilibrium constant expression for each ionization steps is written below.

Explanation:

The chemical formula of carbonic acid is $H_2CO_3$ . It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water