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3 years ago

Please hurry!!! And help!!

Chemistry

1 answer:

pashok25 [27]3 years ago

3 0

Answer:

Gas

Explanation:

This is because a gas is the only substance in the given options which can be expanded or compressed as its particles are far away from each other. They can also take the shapes of the containers they are put in.

HOPE THIS HELPED

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Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %

6 0

3 years ago

How far does light travel in 3.0 days

MArishka [77]

Answer:

$d=7.77\times 10^{13}\ m$

Explanation:

We know that the speed of light is equal to $3\times 10^8\ m/s$

We need to find how far the light travel in 3 days.

Speed of an object is equal to distance covered divided by time taken.

Also, 1 day = 86400 s

3 days = 259200 s

So,

$d=vt$

$d=3\times 10^8\times 259200\\\\d=7.77\times 10^{13}\ m$

So, the light will travel $7.77\times 10^{13}\ m$ in 3 days.

7 0

3 years ago

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

3 0

3 years ago

Electrolysis of molten sodium chloride<br> Products at cathode and anode?<br> Observation?

Mekhanik [1.2K]

At the anode (A), chloride (Cl-) is oxidized to chlorine. ... At the cathode (C), water is diminished to hydroxide and hydrogen gas. The net procedure is the electrolysis of a fluid arrangement of NaCl into mechanically helpful items sodium hydroxide (NaOH) and chlorine gas.

6 0

4 years ago

If the pressure on the surface of water in the liquid stato

Novosadov [1.4K]

The water will boil at C) 80°C

<h3>Further explanation</h3>

Given

Vapour pressure of water = 47 kPa

Required

Boiling point of water

Solution

We can use the Clausius-Clapeyron equation :

$\tt ln(\dfrac{P_1}{P_2})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})$

Vapour pressure of water at boiling point 100°C=101.325 kPa

ΔH vap for water at 100°C=40657 J/mol

R = 8.314 J/mol K

T₁=boiling point of water at 101.325 kPa = 100+273=373 K

Input given values :

$\tt ln\dfrac{101.325}{47}=\dfrac{-40657}{8.314}(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.7682=4890.1852(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.0001571=(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\\dfrac{1}{T_2}=0.00283\rightarrow T_2=353~K=80^oC$

6 0

3 years ago