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3 years ago

Whats the difference between hetergeneous and homogeneous?

1 answer:

4 0

In Homogeneous, "Homo" means same, so homogeneous means the same nature or kind.

Heterogeneous is the opposite. It means diverse.

A good way to remember is just think of homo as same. And the rest should come to you.

Hope this helps! :)

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Rocks and sand are nonliving .all organisms are

solong [7]

Hey there!

Rocks and sand are nonliving. All organisms are living.
Living organisms have five characteristics. Living organisms respond to a stimulus, need energy, grow, reproduce, and get rid of wastes. All living organisms consist of cells.

Hope this helps! Have a Brainly day! :D

5 0

3 years ago

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e

TEA [102]

Given:

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

To determine:

The work done, W

Explanation:

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = -PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ

6 0

3 years ago

Do you really get 7 years of bad luck if you crack a mirror :(

Mazyrski [523]

Answer:

nope its a myth don't worry :)

4 0

3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

2 years ago

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0

3 years ago