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vodomira [7]
3 years ago
9

Which of the following examples depict(s) a physical property?

Chemistry
2 answers:
Allisa [31]3 years ago
7 0

Answer:

\boxed {\tt D. \ All \ of \ the \ above }

Explanation:

A physical property is a property that can be observed without changing the composition.

Some examples of physical properties are:

  • Color
  • Density
  • Length
  • Mass
  • Smell
  • Temperature
  • Boiling Point
  • Hardness

Now, let's examine our answer choices.

Choice A: Sugar is white in color

  • The property is color, which is a physical property.

Choice B: Iron is more dense than aluminum

  • The property is density, which is a physical property.

Choice C: Diamond is harder than graphite.

  • The property is hardness, which is a physical property.

Since all three choices depict physical properties, the best choice is D. All of the above.

dimaraw [331]3 years ago
6 0

Answer:

D

Explanation:

All explain physical no chemical examples were given

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What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)
amid [387]

Answer:

m=59.3gCO_2

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol

Finally, the mass by using the molar mass:

m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2

Best regards.

4 0
2 years ago
Perform an on-line search to discover what conditions are favourable to the growth of moss. be sure to include a comment on ph c
Genrish500 [490]
Moss can grow abundantly in an environment with favorable conditions. They usually grow abundantly in cool, moist and dark places. If not dark, they prefer shady areas. Also, they tend to grow in acidic soil with pH range between 5.0-5.5. 
4 0
3 years ago
s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g
Elenna [48]

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express  your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Now put all the given values in above equation, we get:

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

Hence, the volume of hydrogen gas that will be collected is 1.85 L

8 0
3 years ago
Complete the equation for the conversion of sucrose into glucose <br> (1)C12H22O11 + H2O
Svetlanka [38]

Answer:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Explanation:

Chemical equation:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Source of sucrose:

Sucrose is present in roots of plants and also in fruits. It is storage form of energy. Some insects and bacteria use sucrose as main food. Best example is honeybee which collect sucrose and convert it into honey.

Monomers of sucrose and hydrolysis:

Sucrose consist of monomers glucose and fructose which are join together through glycosidic bond. Hydrolysis break the sucrose molecule into glucose and fructose. In hydrolysis glycosidic bond is break which convert the sucrose into glucose and fructose. Hydrolysis is slow process but this reaction is catalyze by enzyme. The enzyme invertase catalyze this reaction.

The given reaction also completely follow the law of conservation of mass. There are equal number of atoms of elements on both side of chemical equation thus mass remain conserved.

8 0
3 years ago
a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp
Drupady [299]

Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL

Putting values in above equation, we get:

3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M

The concentration of the phosphoric acid solution is 1.172 M

3 0
3 years ago
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