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Copper has a density of 8.96 g/cm3. If 75.0 g of copper is added to 50.0 mL of water in a graduated cylinder, to what volume rea

Tcecarenko [31]

Answer:

The answer to your question is Final volume = 58.37 ml

Explanation:

Data

density = 8.96 g/cm³

mass = 75 g

volume of water = 50 ml

Process

1.- Calculate the volume of copper

Density = mass / volume

Solve for volume

Volume = mass / density

Substitution

Volume = 75/8.96

Simplification

Volume = 8.37cm³ or 8.37 cm³

2.- Calculate the new volume of water in the graduated cylinder

Final volume = 50 + 8.37

Final volume = 58.37 ml

3 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the

qaws [65]

4.1 h = 14760 s

t 1/2 = ln 2 / k 

k = rate reaction = 4.97 x 10^-5 

ln 0.045 / 0.36 = - 4.97 x 10^-5 t 

2.08 = 4.97 x 10^-5 t 

t = 41839.9 s = 11 h 37 min 19 s

3 0

3 years ago

which of the following will stay constant,no matter is the substance is in a solid,liquid,or gas states

Vlada [557]

I believe it is a gas

6 0

3 years ago

Read 2 more answers

For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0

3 years ago