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3 years ago

A) An iron nail is attracted to a magnet.

Chemistry

2 answers:

11111nata11111 [884]3 years ago

4 0

there's no question on here

QveST [7]3 years ago

3 0

5°C is the answer I think but idk for sure

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3. People and instruments produce different ranges of sound temperatures is this true or false

anzhelika [568]

True I believe since every person has a different temperature and sound waves.

6 0

3 years ago

A plant grows into a shrub a physical or chemical change?<br>Explain why?

kherson [118]

A plant growing into a shrub is an example of a chemical change because once the plant is in its shrub form you can't revert it back.

8 0

3 years ago

Hydrochloric acid is widely used as a laboratory reagent, in refining ore for the production of tin and tantalum, and as a catal

GaryK [48]

Answer:

Option 4

Explanation:

Hydrochloric acid is a strong one, that gives protons to medium. It can be dissociated as this:

HCl → H⁺ + Cl⁻

M means Molarity. It is a sort of concentration that indicates the moles of solute in 1L of solution.

M = moles / volume (L)

We can also say M = mmoles / mL of solution

M . mL = mmoles

0.453 M . 62.85mL = 28.5 mmoles

If we divide by 1000 → 28.5 mmol . 1 mol / 1000 mmol = 0.0285

3 0

3 years ago

How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(

Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
O₂: 2 moles
CO₂: 1 mole
H₂O: 2 moles

The molar mass of the compounds is:

CH₄: 16 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
O₂: 2 moles ×32 g/mole= 64 grams
CO₂: 1 mole ×44 g/mole= 44 grams
H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

$mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}$

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

$mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}$