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3 years ago

20 POINTS! what would be an example of an everyday household acid and base?

Chemistry

2 answers:

Elza [17]3 years ago

4 0

Answer:

Baking Soda, Vinegar, Ammonia

Explanation:

Bogdan [553]3 years ago

4 0

vinegar is a household acid and washing soda is a very strong base

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Melting can be best described as a process in which molecules

statuscvo [17]

Melting can be best described as a process in which molecules gain enough kinetic energy to be able to pass to each other.

3 0

3 years ago

A 365 L balloon is blown up inside at 283 K. It is then taken outside in the hot sun and Heated to 300K. What is the new volume?

asambeis [7]

V₁ = initial Volume of the balloon after it is blown up = 365 L

V₂ = new Volume of the balloon after it is taken outside = ?

T₁ = initial temperature of the balloon = 283 K

T₂ = new temperature of the balloon = 300 K

using the equation

V₁/V₂ = T₁/T₂

365/V₂ = 283/300

V₂ = 387 L

6 0

4 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

Juli2301 [7.4K]

Answer:

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

Explanation:

The gas ideal law is

PV= nRT (equation 1)

Where:

P = pressure

R = gas constant

T = temperature

n= moles of substance

V = volume

Working with equation 1 we can get

$n =\frac{PV}{RT}$

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

$\frac{m}{mw} =\frac{PV}{RT}$ or

$m =\frac{P*V*mw}{R*T}$ (equation 2)

The cylindrical container has a constant pressure p

The volume is the volume of a cylinder this is

$V =(pi)*r^{2}*h$

Where:

r = radius

h = height

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is $V =(pi)*r^{2}*h$

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:

$T =\frac{T_{1} + T_{O}}{2}$

Replacing these values in the equation 2 we get:

$m =\frac{P*V*mw}{R*T}$ (equation 2)

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

8 0

3 years ago

given two equations representing reactions: which type of reaction is represented by each of these equations?

Sergeeva-Olga [200]

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

7 0

3 years ago