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2 years ago

Which statement best explains why heating a liquid affects its velocity

1 answer:

3 0

Part of the answer you are looking for may be- Molecules overcome attractions easily and at high temperatures, move faster- (paraphrased from Google) Since the question asked was not provided with the statements as proposed in the original question, I cannot give you an exact answer. I did what i could do, hope this helps!

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How many grams of H2O will be formed when 36.8 g H2 is mixed with 40.2 g O2 and allowed to completely react to form water

Artemon [7]

Answer:

45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂

Explanation:

The balanced reaction is:

2 H₂ + O₂ → 2 H₂O

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

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

Being the molar mass of the compounds:

H₂: 2 g/mole
O₂: 32 g/mole
H₂O: 18 g/mole

then, by reaction stoichiometry, the following amounts of reactant and product mass participate:

H₂: 2 moles* 2 g/mole= 4 g
O₂: 1 mole* 32 g/mole= 32 g
H₂O: 2 moles* 18 g/mole= 36 g

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.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 4 g of H₂ react with 32 g of O₂, 36.8 g of H₂ with how much mass of O₂ will it react?

$mass of O_{2} =\frac{36.8 grams of H_{2}*32 grams of O_{2} }{4 grams of H_{2}}$

mass of O₂=294.4 grams

But 294.4 grams of O₂ are not available, 40.2 grams are available. Since you have less mass than you need to react with 36.8 grams of H₂, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry 32 grams of O₂ form 36 grams of H₂O, 40.2 grams of O₂ how much mass of H₂O will it form?

$mass of H_{2}O=\frac{40.2 grams of O_{2} *36 grams of H_{2}O }{32 grams of O_{2} }$

mass of H₂O= 45.225 grams

<u><em>45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂</em></u>

8 0

2 years ago

PLEASE HELPPPPPPP:

crimeas [40]

Answer:

part 1 is from the article

Explanation:

5 0

3 years ago

Why is copper used to make electric wires?

sesenic [268]

Question:

Copper is used in a variety of applications, one of which is electrical wires. Whether it’s a power cable for a TV, computer, kitchen appliance or construction tool, it’s probably made of copper. Featuring the atomic number 29, this red-brown metal has become essential to our everyday life by helping power countless devices. But copper isn’t the only material that can transmit electricity, so why is it preferred for creating electrical wires?

Answer:

<h2>The two most important factors on why Copper is used to make not just electric wires, but lots of electric related things is that they are Inexpensive and Super Conductive</h2>

Explanation:

Now, I will explain the reasons why these are good things about the copper wire.

<h2>High Conductivity </h2>

Surpassed only by silver, copper is a highly conductive metal. This means electricity can pass through it with greater ease, making it ideal for use in electrical wires. Companies can use other conductive metals to create electrical wires. Unless they use silver, though, the high conductivity properties of copper allow for a greater distance of electrical current travel. Companies can create longer, better performing electrical wires using copper instead of most other conductive metals.

<h2>Inexpensive </h2>

Copper is also relatively inexpensive when compared to other metals. Gold, for example, is an excellent conductor of electricity, but it costs several times more than copper. If companies used gold to manufacture electrical wires, they’d essentially waste money, as copper is more conductive and costs less than its goal counterpart. This alone is reason enough to make copper the de-facto standard for electrical wires.

<h2>High Ductility </h2>

Copper isn’t just conductive; it’s also ductile. In other words, you can bend and flex copper — to some degree — without it breaking or otherwise sustaining damage. Why is this important? Well, electrical wires must often travel through walls, floors, ceilings and other tight spaces. As a result, they’ll naturally bend and snake their way around the home or building in which they are used. The ductile properties of copper allow copper electrical wires to bend and flex. They’ll still transmit electricity, and they won’t lose any power strength from shape deformity.

<h2>Thermal Resistant </h2>

A benefit of copper electrical wires that’s often overlooked is its thermal-resistant properties. According to ESFI, electrical fires are responsible for more than 51,000 residential house fires in the United States each year — and that’s not accounting for commercial/business fires. Copper electrical wires are safer to use than wires made of most other conductive metals because they are resistant to heat.

<h2>As you can see, copper is the preferred metal for electrical wires for several reasons. It has high electrical conductive; it’s inexpensive; it’s ductile; and it’s thermal resistant. This makes is the universally accepted standard for the manufacturing of electrical wires.</h2>

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3 0

2 years ago

Part IV. Limiting Reactants! A Challenge Problem!

Alexxandr [17]

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles). Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %

3 0

2 years ago

Which sample contains more atoms:<br> 3.89 g of nickel or 6.61 g of silver?<br> Defend your answer.

Alenkinab [10]

Nickel contains more atoms

3 0

2 years ago