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lorasvet [3.4K]
3 years ago
7

8th grade science

Chemistry
2 answers:
eimsori [14]3 years ago
7 0

Answer:

B

Explanation:

morpeh [17]3 years ago
4 0

Answer:

Copper and Iron

Explanation:

Metals, like copper and iron, are good conductors of heat; that is because  metals are bounded by a 'cloud' of free electrons and these electrons can carry easily kinetic energy through the metal

On the other hand, foam and wood are isolators, that is, they are bad conductors of heat.

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When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c
PolarNik [594]

Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 *  3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

3 0
3 years ago
List 4 things that can affect the rate of reaction during a chemical reaction.
laiz [17]

Answer:

Reactant concentration, the physical state of the reactants, and surface area, temperature, and the presence of a catalyst are the four main factors that affect reaction rate.

Explanation:

6 0
3 years ago
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
A student wants to remove the salt from a mixture of sand and salt in order to get only pure sand. He adds water to the mixture.
raketka [301]

Answer:

Here is one way: Add water to the mixture. Only the sugar dissolves. This is a physical change.

Explanation:

The sugar would dissolve in water. You could then pour off the solution and wash the remaining sand with a bit more water. Heat the water to evaporate it from the sugar, and the two are separated.

5 0
3 years ago
Read 2 more answers
If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?
krok68 [10]

Answer:

\large\boxed{\large\boxed{0.64M}}

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

          \text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}

Substitute and solve for the molarity of the diluted solution:

           250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M

8 0
3 years ago
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