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1 year ago

Which are examples of equilibrium? Check all that apply.

Chemistry

2 answers:

Maslowich1 year ago

5 0

Answer:

B and C.

Explanation:

The equilibrium means that amount that are coming equals the amount that are going.

B. A person's income and expenses are equal every month.

C. The pressure exerted on the inside of a balloon by the gas inside is equal to the pressure exerted on the outside of the balloon by the atmosphere.

Sveta_85 [38]1 year ago

4 0

I believe that the answers are 3 and 4

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We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find

iragen [17]

Explanation:

a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:

$V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V$

Gold $_{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}$

$1 \mu=10^{-6} \mathrm{m}$

First, find the volume of the sample and then find the area of the sample.

$V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3}$ $=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}$

$V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}$

b. Using the provided information from part $a$ ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

$V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}$

$h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}$

8 0

2 years ago

The general term for a large molecule made up of many similar subunits is

Setler79 [48]

A) Polymer is the general name of large units made of many smaller units (these would be called monomers). An example is starch, this is a carbohydrate polymer that is made up of smaller units (monomers) called glucose.

6 0

2 years ago

At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

7 0

2 years ago

25.0 ml of a 6.0 m hno3 stock solution is diluted using water to 100 ml. How many moles of hno3 are present in the dilute soluti

lana66690 [7]

Answer:

The answer to your question is: 6 moles of HNO₃

Explanation:

Data

Volume = 25 ml

Concentration = 6 M HNO₃

Diluted 100 ml

Formula

Molarity = # moles / volume

# of moles = Volume x Molarity

Process

# of moles = 0.10 x 6

= 6 moles

6 0

2 years ago

Read 2 more answers

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

Galina-37 [17]

3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4

3 0

1 year ago