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3 years ago

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A given atom has 3 protons, 4 neutrons, and 3 electrons. Describe how you can use those three values to describe the structure,

charge, and mass of the atom.

1 answer:

ELEN [110]3 years ago

3 0

Charge # = protons - electons

Mass # = protons + neutrons

so that would be

3-3= charge#
3+4= mass#

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A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled

marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so that the concept could be applied

Initial volume of the AICI3 solution $=23.1 \mathrm{mL}$

Initial Molarity of the solution $=833 \mathrm{mM}$

Final molarity of the solution $=75.0 \mathrm{mM}$

Final volume of the solution $=?$

From Law of Dilution, $M_{f} V_{f}=M_{i} V_{i}$

$\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L$

Final Volume of the solution $=0.257$

7 0

3 years ago

Use the following equation to answer the questions and please show all work.

DIA [1.3K]

Answer:

a.36 g of water is produced.

b.64 g of $O_{2}$ is consumed.

Explanation:

The reaction is $2H_{2} + O_{2}$ ⇒ $2H_{2}O$

a.

Given,

Weight of $H_{2}$ reacted = 4g

Weight of 1 mole of $H_{2}$ = 2 $\times$ 1 = 2g

Therefore no. of moles of $H_{2}$ reacted = $\frac{4}{2}$ = 2 moles;

Also given,

Weight of $O_{2}$ reacted = 32 g

Weight of 1 mole of $O_{2}$ = 2 $\times$ 16 = 32 g

Therefore no. of moles of $O_{2}$ reacted = $\frac{32}{32}$ = 1

We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,

As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of $H_{2} O$ produced = 2

Weight of 1 mole of water = $2\times 1+16$ = 18

Therefore weight of $H_{2}O$ produced = $2\times 18$ = 36gm

b.

Given ,

72 g of $H_{2}O$ is produced.

So,

no. of moles of $H_{2}O$ produced = $\frac{72}{18}$ = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of $O_{2}$ reacted = $2\times32$ = 64 g

Method 2:

Given,

8 g of $H_{2}$ has reacted.

So,

no. of moles of $H_{2}$ reacted = $\frac{8}{2}$ = 4 moles.

From equation , we know that For every 2 moles of $H_{2}$ reacted,1 mole of $O_{2}$ will react.

Therefore,

No. of moles of $O_{2}$ that reacts with 4 moles of $H_{2}$ = $2\times1$ = 2 moles

Therefore the weight of $O_{2}$ reacted = $2\times 32$ = 64 g

6 0

3 years ago

Given the balanced equation representing a reaction: 2h2o(l) + 571.6 kJ -> 2h2(g) + o2(g). What occurred as a result of this

Ivan

Answer: option D) energy was absorbed and entropy increased.

Explanation:

1) Given balanced equation:

2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).

2) Being the energy placed on the side of the reactants means that the energy is used (consumed or absorbed). This is an endothermic reaction.

So, the first part is that energy was absorbed.

3) As for the entropy, it is a measure of the disorder or radomness of the system.

Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.

With that, you have shown that the right option is D) energy was absorbed and increased.

7 0

4 years ago

Thisss for you blake :)))

Lady_Fox [76]

I’m not Blake but ok

4 0

3 years ago

Read 2 more answers

The half-life of cobalt-60 is 5.26 years. If a sample originally contained 220 grams of this isotope, how many grams would be le

ser-zykov [4K]

Answer:

Question Answer

The half-life of cobalt-60 is 5.26 years. How many half-lives have passed in 10.52 years? 2

12.5% of a radioactive sample are left. How many half-lives have passed? 3

After 3 half-lives, how much of a 400 gram sample of radioactive uranium remains? 50g

Explanation:

7 0

3 years ago