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vesna_86 [32]
3 years ago
10

How many moles of copper are needed to react with sulfur to produce 0.25 moles Ol

Chemistry
1 answer:
Tems11 [23]3 years ago
6 0

Answer:

                       0.50 moles of Cu

Explanation:

                    The balanced chemical equation for given synthetic reaction is,

                                          2 Cu + S → Cu₂S

According to balance chemical equation,

                  1 mole of Cu₂S is produced by  =  2 moles of Cu

So,

          0.25 moles of Cu₂S will be produced by  =  X moles of Cu

Solving for X,

                      X  =  0.25 mol × 2 mol /  1 mol

                      X  =  0.50 moles of Cu

Hence, as the molar ratio of Cu to Cu₂S is 2:1 hence, to produce 0.25 moles of Cu₂S we will need 0.50 moles of Cu.

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D water expands when it freezes

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What is the total pressure exerted by a mixture of 48.0 grams of CH4 and 56.0 grams of
Oksi-84 [34.3K]

The pressure of the gas is obtained as 48 atm.

<h3>What is the total pressure?</h3>

Now we know that;

Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles

Number of moles of H2 =  56.0 grams/2 g/mol = 28 moles

Total number of moles present = 3 moles + 28 moles = 31 moles

Using;

PV =nRT

P = total pressure

V = total volume

n = total number of moles

R = gas constant

T = temperature

P = nRT/V

P = 31 * 0.082 * 286/15

P = 48 atm

Learn more about pressure of a gas:brainly.com/question/18124975

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4 0
1 year ago
Read the sentence from the introduction [paragraphs 1-3].
ss7ja [257]

Answer: the answer is B

Explanation:

6 0
3 years ago
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
match the number of negative charges to the number of positive charges to make each combination balanced (see image for answer)
BartSMP [9]

Answer : The correct match is:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

Explanation :

As we now that there are three subatomic particles which are protons, electrons and neutrons.

The protons and neutrons are located inside the nucleus and electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and neutrons are neutral.

As we know that all the things are made up of charges and opposite charges attract to each other.

In a neutral atom, the positive charges and negative charges are balanced in an object. That means, in neutral atom the number of positive charges are equal to the negative charges.

So we can say that:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

5 0
3 years ago
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