If two oxygen atoms bond covalently, the bond is

This element exists in adundance in the sun.Explain how you would go about capturing sunlight.Would this captured sunlight conta

nasty-shy [4]

Yes because the sun will make alots of thing grow

4 0

3 years ago

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/

Ivenika [448]

Answer:

The answer to your question is: C₃H₃O This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

44 g -------------------- 12 g

3.2007 g ------------ x

x = (3.2007 x 12) / 44

x = 0.8729 g of Carbon

12 g of C -------------- 1 mol

0.8729 g -------------- x

x = (0.8729 x 1) / 12

x = 0.0727 mol of Carbon

For Hydrogen

18 g ---------------------- 1 g

1.3102 g ------------------- x

x = (1.3102 x 1) / 18

x = 0.0727 g of Hydrogen

1 g ------------------------ 1 mol

0.0727g ---------------- x

x = (0.0727 x 1)/1

x = 0.0727 mol of Hydrogen

For oxygen

g of Oxygen = g of sample - g of Carbon - g of hydrogen

g of Oxygen = 1.3109 - 0.8709 - 0.0727

g of Oxygen = 0.3673

16 g of Oxygen ------------- 1 mol of O

0.3673 g --------------------- x

x = (0.3673 x 1)/ 16

x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon 0.0727 / 0.023 = 3.1 ≈ 3

Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3

Oxygen 0.0230 / 0.023 = 1

C₃H₃O

8 0

3 years ago

Read 2 more answers

How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?

nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

Concentration of the concentrated solution (C₁): 2 M

Volume of the concentrated solution (V₁): ?

Concentration of the diluted solution (C₂): 0.1 M

Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

4 0

3 years ago

15.0 g of Fe and 25.0 g of sand are added to 250.0 g of water. a. Determine the percent mass of Fe, sand, and water in the mixtu

lana [24]

Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand = 8.62%

percentage mass of water = 86.205%

B. (Iron + sand + water) -------> ( iron + sand) ------> sand

C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

percentage mass of iron = 15/290 * 100% = 5.17%

percentage mass of sand = 25/290 * 100% = 8.62%

percentage mass of water = 250/290 * 100% = 86.205%

B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

5 0

3 years ago

Fe(s) + O2 (g) → Fe3+ O2− What most likely happens during this reaction?

Musya8 [376]

Answer : The most likely happens during this reaction is, Oxidation-reduction

Explanation :

The balanced reaction will be,

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

In this reaction, neutral iron loses 3 electrons and oxidizes in (+3) state, $Fe^{3+}$ and neutral oxygen gains 2 electrons and reduces in (-2) state, $O^{2-}$

When iron react with oxygen gas to give iron oxide. This process is known as iron rusting. During the reaction, oxidation-reduction process occurs.

Oxidation : It is a type of chemical reaction in which a substance loses its electrons. Or we can say that in oxidation, the oxidation number increases.

Reduction : It is a type of chemical reaction in which a substance gains its electrons. Or we can say that in reduction, the oxidation number decreases.

5 0

3 years ago