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Rina8888 [55]
3 years ago
13

PLEASE ANSWER THIS FAST!! (I’ll give ALL my points)!!

Chemistry
1 answer:
Lapatulllka [165]3 years ago
4 0

Answer:

Explanation:

hakuna

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Why do we perform an experiment?
7nadin3 [17]

Answer:

Experiment plays many roles in science. One of its important roles is to test theories and to provide the basis for scientific knowledge. It can also call for a new theory, either by showing that an accepted theory is incorrect, or by exhibiting a new phenomenon that is in need of explanation.

Explanation:

6 0
3 years ago
Read 2 more answers
To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7
Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0
2 years ago
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. when 2.59 g of magnesium ribbon bu
stepan [7]
Burning Mg in the air and reacting with O2 forming a white powder of MnO

So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)

to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
                        Δ
2Mg(s) + O2(g)⇒ 2MgO(s)

 

4 0
3 years ago
In the reaction shown above , the?
matrenka [14]

Answer:

I think option (d) is right answer

8 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
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