Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.

Answers

Answer 1

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

Hence the empirical formula is CHO

Answer 2

Answer:

The empirical formula of X is C3H4O3.

Explanation:


Related Questions

2 2 6 2 6 2 10 3
1s 2s 2p 3s 3p 4s 3d 4p
=

Answers

Answer:

ARSENIC

Explanation:

It has an atomic number of 33

If the earth was a guava fruit, the space where the seeds are would be the core/mantle​

Answers

Right on ! I need to answer a question to get mine answered so here I am :)

Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. What is most likely the mass of the blue copper sulfate that he heated and why?

Answers

Answer: The mass of blue copper sulfate is 3.5 g

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the heating of copper sulfate crystals is:

Let the mass of blue copper sulfate be 'x' grams

We are given:

Mass of copper sulfate powder = 2.1 grams

Mass of water = 1.4 grams

Total mass on reactant side = x

Total mass on product side = (2.1 + 1.4) g

So, by applying law of conservation of mass, we get:

Hence, the mass of blue copper sulfate is 3.5 grams

In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.

Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?

Answers

Answer:

a) - 0.2 M

b) - 0.2 M

c)- 0

Explanation:

The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:

MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol

a). Molarity = moles CuBr₂/1 L solution

moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol

Volume in L = 375 mL x 1 L/1000 mL = 0.375 L

M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M

b). When is added to water, CuBr₂ dissociates into ions as follows:

CuBr₂ ⇒ Cu²⁺ + 2 Br⁻

We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:

0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M

c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).

A balloon contains 1.1 L of gas at a pressure of 0.80 atm. How will the volume
change if the pressure is increased to 2.0 atm?

Answers

Answer:

Final volume  = 0.44 L

Explanation:

Given data:

Initial volume of balloon = 1.1 L

Initial pressure = 0.80 atm

Final volume = ?

Final pressure = 2.0 atm

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

0.80 atm × 1.1 L = 2.0 atm × V₂

V₂ = 0.88 atm. L/ 2.0 atm

V₂ = 0.44 L

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with potassium hydroxide. It requires 22.9 mL of 1.430 M KOH solution to titrate both acidic protons in 54.2 mL of the carbonic acid solution.

Required:
a. Write a balanced net ionic equation for the neutralization reaction. Include physical states.
b. Calculate the molarity of the carbonic acid solution.

Answers

Answer:

a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

b. 0.603 M

Explanation:

Step 1: Write the neutralization reaction

H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

Step 2: Calculate the reacting moles of KOH

22.9 mL of 1.430 M KOH react.

0.0229 L × (1.430 mol/L) = 0.0327 mol

Step 3: Calculate the reacting moles of H₂CO₃

The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.

Step 4: Calculate the molarity of H₂CO₃

0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:

M = 0.0327 mol/0.0542 L = 0.603 M

Problem:
[Ar]4s2
Identify the period (p) , group (g) and valence electrons block of the element

Answers

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.

Answers

Answer:

QC= [O2]^3/[F2]^10

Explanation:


1. What 2 subatomic particles have charges? List the particle name and its charge.

Answers

Answer: Proton - positive charge (+)

Neutron - neutral charge (0)

Electron - negative charge (-)

Explanation:

A student measured the masses of four different-sized blocks. The student determined that each block had a mass of 50 grams.


(There is a small block, a little bit bigger block, a big block and the biggles block)


Which block has the least density?

Answers

Answer:..

Explanation:

Which type of reaction has the general formula of AB + CD → AD + CB?

synthesis
decomposition
oxidation
replacement

Answers

Answer:

The answer is D. Replacement

The given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

The given general reaction has been:

[tex]\rm AB\;+\;CD\;\rightarrow\;AD\;+\;CB[/tex]

In the given reaction, there has presence of compound AB and CD. The reaction between the two results in the formulation of AD and CB. In the reaction, the B has been more attracted by C and forms bond with C by replacing D.

In the same way, A has been more attracted with D and forms bond with D replacing C.

The type of reaction in which the more reactive element replaces the less reactive element has been termed as replacement reaction. Thus, the given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

For more information about replacement reaction, refer to the link:

https://brainly.com/question/8625202

The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga

Answers

Answer:

2.35 x 10²⁰ atoms Ga

Explanation:

After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.

[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]

Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.

The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

Stoichiometry

From the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.

First, we will determine the number of moles of Ga present

Using the formula,

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]

Mass = 27.2 mg = 0.0272 g

Molar mass = 69.72 g/mol

Then,

[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]

[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles

Now, for the number of atoms present

From the formula

Number of atoms = Number of moles × Avogadro's constant

Then,

Number of Ga atoms = 0.000390132 × 6.022×10²³

Number of Ga atoms = 2.35 × 10²⁰ atoms

Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

Learn more on stoichiometry here: https://brainly.com/question/14464650

plz help answer both will mark brainest

Answers

1st one is balanced other one i believe is not
hope that helps:)

SOMEONE PLZ HELP!!!!

Answers

Answer:

4.22mL

Explanation:

V=m/d

v= 18.45g/4.37g/mL

A certain chemical reaction releases of heat for each gram of reactant consumed. How can you calculate the heat produced by the consumption of of reactant? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.

Answers

Complete Question

The complete question is shown in the first uploaded image

Answer:

So the math expression is  

             [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

Explanation:

From the question we are told that

  The heat released for 1 gram of reactant consumed is  [tex]H  =  37.5 \ KJ/g [/tex]

   The mass of reactant considered is  [tex]m =  1.9 \ kg  =  1900 \  g[/tex]

So  if

             [tex]37.5 \ KJ [/tex] is produced for  1 gram

Then

              x kJ is produced for  1900 g  

=>   [tex]x  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

So the heat released is  

       [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

Which is one way that minerals crystallize from materials dissolved in water?

from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil

Answers

Answer:

the second answer its science behind it

Answer:

b

Explanation:

Help, 8th grade Science

Answers

It’s for sure control container

PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things?
organic

inorganic

acidic

nonacidic

Answers

Answer:

acidic because of electrical issues and the body of electrical equipment

4. Horizontal rows of the Periodic Table are called:
a, Clusters
Groups
b. Families
d) Periods

Answers

The horizontal rows in the periodic table are periods, while the vertical rows are called groups

If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction

Answers

Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]

= 2.35 × 10⁻⁴ mol

Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]

Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]

Δ[O₂]  = 0.04626 M

The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

= [tex]\dfrac{0.04626}{60}[/tex]

= 7.71 × 10⁻⁴ M/s

What is the pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution

Answers

Answer:

pH = 11

Explanation:

The equilibrium of a weak base as NaCN in water is:

NaCN(aq) + H₂O(l) ⇄ OH⁻(aq) + Na⁺(aq) + HCN(aq)

And kb, the equilibrium constant, is:

Kb = [OH⁻] [HCN] / [NaCN]

Where Kb of NaCN is 2.04x10⁻⁵

In the beginning, the [NaCN] is 0.050mol / L = 0.050M.

Both [OH⁻] and [HCN] are produced from this equilibrium, and its concentration is X, that is:

2.04x10⁻⁵ = [X] [X] / [0.050M]

1.02x10⁻⁶ = X²

X = 1x10⁻³ = [OH⁻]

As pOH = - log [OH⁻]

pOH = 3.00

And pH = 14 - pOH

pH = 11

What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.

Answers

Answer:

i

[tex]J_{m} = 20 [/tex]

ii

[tex]J_{m} = 22.5 [/tex]

Explanation:

From the question we are told that

  The first temperatures is [tex]T_1 =  25^oC =  25 +273 =298 \ K[/tex]

   The second temperature is  [tex]T_2 =  100^oC =  100 +273 = 373 \ K[/tex]

Generally the equation for  the most highly populated rotational energy level is mathematically represented as

     [tex]J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}[/tex]

Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]

Also  

      B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]

   Generally the energy require per mole to move 1 cm is  12 J /mole

So   [tex]0.244 \ cm^{-1}[/tex]  will require x J/mole

           [tex]x =  0.244 *  12[/tex]

=>          [tex]x =  2.928 \ J/mol [/tex]

So at the first temperature

     [tex]J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 20 [/tex]

So at the second temperature

           [tex]J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 22.5 [/tex]

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