All categories

3 years ago

What is heterogenous mixture

Chemistry

1 answer:

Kamila [148]3 years ago

7 0

Answer:

A heterogeneous mixture is simply any mixture that is not uniform in composition - it's a non-uniform mixture of smaller constituent parts. By contrast, a mixture that is uniform in composition is a homogeneous mixture.

You might be interested in

what is the specific heat of a 25.0 g substance that absorbs 493.4 j and raises the temperature from 12.0°c to 34°c?

stellarik [79]

Answer:

Specific heat of substance = 0.897 J/g.°C

Explanation:

Given data:

Specific heat of substance = ?

Mass of substance = 25.0 g

Heat absorbed = 493.4 J

Initial temperature = 12.0 °C

Final temperature = 34°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 34°C -12.0°C

ΔT = 22°C

493.4 J = 25 g ×c× 22°C

493.4 J = 550 g.°C ×c

c = 493.4 J / 550 g.°C

c = 0.897 J/g.°C

4 0

3 years ago

What is a low pressure system?

KonstantinChe [14]

A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.

<u>Explanation</u>:

Wind flow towards the low pressure and the air rises in the atmosphere. As the air increases, the water vapor within it solidifies, forming clouds and undergo precipitation. Low pressure formed in the center areas.

The atmospheric circulations of air up and down in a low-pressure area remove a small amount of atmosphere. This usually happens between warm and cold air masses by flowing air which tries to reduce the contrast of temperature.

5 0

3 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago

Tornadoes can be identified and tracked using advanced instruments that can detect small changes in wind velocity and air pressu

sdas [7]

C would be the only logical answer if you really think about what it’s saying about the instrument.

6 0

4 years ago

What are the two half-reactions that occur in this electroplating process. Be sure to include the states of the elements or ions

Serhud [2]

<span>This is rather a case of purification of impure copper or extraction of Cu from its alloy. You need to place the impure copper rod on the positive electrode (Anode) usualyy made of carbon rod, whereby oxidation reaction takes place: Cu (s) -------> Cu2+ (aq) + 2e- That is to say the impure cooper dissolves into solution. The copper (II) ions move to the negative electrode (cathode) usually made of pure copper rod. At the cathode, the Cu2+ ions are reduced : Cu2 (aq)+ + 2e- -----> Cu(s). That is to say the copper (II) ions are deposited as solid copper atoms onto the cu-rod electrode. In this way impure copper is deposited as pure copper onto the copper cathode</span>

4 0

4 years ago