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3 years ago

Describe how the molecular models you assembled are similar to real molecules.

Chemistry

1 answer:

mario62 [17]3 years ago

4 0

Answer:

Atoms models are bigger but they don not match the true or positive way of atoms. to make them visible scientists make their models that resembled with the real ones.

Explanation:

Molecules are microscopic entities and only can be seen under a microscope.

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What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0

4 years ago

A sample of methane gas having a volume of 2.80 L at 25 degree C and 1.65 atm was mixed with a sample of oxygen gas having a vol

Svetradugi [14.3K]

Answer:

Volume of carbon dioxide = 2.47 L

Explanation:

Given that:

For methane gas:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25.0 + 273.15) K = 298.15 K

V = 2.80 L

Pressure = 1.65 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1.65 atm × 2.80 L = n ×0.0821 L atm/ K mol × 298.15 K

⇒n for methane gas = 0.1887 mol

For oxygen gas:-

Temperature = 31 °C

T = (31 + 273.15) K = 304.15 K

V = 35.0 L

Pressure = 1.25 atm

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

1.25 atm × 35.0 L = n ×0.0821 L atm/ K mol × 304.15 K

⇒n for oxygen gas = 1.7521 mol

According to the reaction:-

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

Also,

0.1887 mole of methane gas reacts with 2*0.1887 moles of oxygen gas

Moles of oxygen gas = 0.3774 moles

Available moles of oxygen gas = 1.7521 moles

Limiting reagent is the one which is present in small amount. Thus, methane gas is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction forms 1 mole of carbon dioxide

Thus,

0.1887 mole of methane gas on reaction forms 0.1887 mole of carbon dioxide

Moles of carbon dioxide = 0.1887 moles

Given that:- Pressure = 2.50 atm

Temperature = 125 °C

T = (125 + 273.15) K = 398.15 K

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

2.50 atm × V = 0.1887 moles ×0.0821 L atm/ K mol × 398.15 K

Volume of carbon dioxide = 2.47 L

7 0

3 years ago

In Kevin’s favorite comic book, the superhero has special powers. In one adventure, the superhero must stop an evil villain who

KATRIN_1 [288]

Answer:

Explanation:

The law of conservation of energy states that energy cannot just be created, so answer one is out. Energy also cannot be simply destroyed, so answer to is out as well. From here it gets tricky, since both answers make some sense under these rules. However, absorbed is bassically the same as destroying it leaving only answer three. Answer three does obey these laws, as energy can be converted.

3 0

2 years ago

The carbon atom has a total of _________ electrons. A.6 B.4 C.2 D.8

denis23 [38]

Carbon has A. 6 atoms

5 0

3 years ago

Read 2 more answers

This is craeted when white light is passed throught a prism

dlinn [17]

When light is passed through a prism it refracts and reflects i just did this in school hope it helped

3 0

3 years ago