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3 years ago

What happens when the two atoms get even closer?

Chemistry

2 answers:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

If two atoms get close enough together then the electrons of each atom will be attracted to both nuclii

Explanation:

just olya [345]3 years ago

5 0

Answer:

If two atoms get close together then the electrons of each atom will get attracted to the other one and bc one atom lose electrons gets a positive charge and the one with more electrones will get a negetive charge

Explanation:

brailiest pls pls pls

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How many moles of HCl are present in 40.0 mL of a 0.035 M solution? 1. 0.0014 mol 2. 0.0060 mol 3. 0.25 mol?

Vinil7 [7]

40.0mL(1 L/1000 mL) = 0.040 L

then plug into the formula M = moles/liters 

0.035 M = moles/ 0.040L 

multipy both sides by 0.040L, and you get 0.0014 moles 

so the answer is 1

4 0

3 years ago

If 125g of KClO3 is heated, what is the total mass of the products?

Andrej [43]

Given parameters:

Mass of KClO₃ = 125g

Unknown:

Total mass of the products = ?

When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

2KClO₃ → 2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

Find the number of moles of the given reactant:

Number of moles = $\frac{mass}{molar mass}$

molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

So number of moles of KClO₃ = $\frac{125}{122.5}$ = 1.02moles

2. Now, using this number of moles, find the number of moles of the products using this value;

2 moles of KClO₃ produced 2 moles of KCl

1.02 moles of KClO₃ will also produce 1.02moles of KCl

2 moles of KClO₃ produced 3 moles of O₂

1.02 moles of KClO₃ will produce $\frac{1.02 x 3} {2}$ mole = 1.53 moles of O₂

3. Now find the masses of each product;

Mass = number of moles x molar mass

molar mass of KCl = 39 + 35.5 = 74.5g/mol

molar mass of O₂ = 16 x 2 = 32g/mol

Mass of KCl = 74.5 x 1.02 = 75.99g

Mass of O₂ = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

= 124.95g

This value is approximately the same as that of mass of KClO₃

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3 years ago

This is for chemistry:/

Andreyy89

Answer:

Its the bottom one :) ......

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3 years ago

In the science fiction movie, The Abyss, a diver is able to breathe while his head is immersed in a specially prepared liquid pu

Klio2033 [76]

Answer:

Explanation:

This type of experiment was carried out in 1960s on rodents, it was partially successful but was perceived impractical and dangerous for humans,it is possible theoretically.

Oxygen is broken down or dissolves in a thin film of fluid in the alveoli, surprisingly in normal breathing liquid composed of dissolved oxygen is involved. Evidently respiratory gas must be able to dissolve in this liquid and in concentration required to keep the partial pressure necessary to power diffusion.

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3 years ago

What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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3 years ago