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2 years ago

Which on goes where please help!!!

Chemistry

1 answer:

EastWind [94]2 years ago

6 0

Pretty easy.. The ice is a solid, the rain is a liquid the drink is a liquid, the water drain is a liquid, the wood is a solid,

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1 2 3 4 5 6 7 8 9 10TIME REMAINING53:56A scientist told his student assistants that he had sorted groups of mice by their intell

Gekata [30.6K]

Answer:

People can introduce their own biases into an experiment.

Explanation:

4 0

2 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

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 = $4.0\times 10^1$

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

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

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

Please help ASAP

DochEvi [55]

Aluminum. chemical element with symbol Al

4 0

3 years ago

The nonvolatile, nonelectrolyte DDT , C14H9Cl5 (354.5 g/mol), is soluble in benzene, C6H6. Calculate the osmotic pressure (in at

Ainat [17]

Step 1

The osmotic pressure is calculated as follows:

$\begin{gathered} \pi\text{ = C x R x T} \\ C\text{ = molarity = }\frac{moles\text{ of solute}}{Volume\text{ of solution \lparen L\rparen}} \\ R\text{ = 0.082 }\frac{atmxL}{mol\text{ x K}} \end{gathered}$

-------------

Step 2

Information provided:

The mass of solute = 13.6 g

Volume of solution = 251 mL

Absolute temperature = T = 298 K

The molar mass of solute = M = 354.5 g/mol

-------------

Step 3

Procedure:

1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L

---

C = moles of solute/volume of solution (L)

C = mass of solute/(molar mass x Volume (L))

C = 13.6 g/(354.5 g/mol x 0.251 L)

C = 0.153 mol/L

---

π = C x R x T

π = 0.153 mol/L x 0.082 atm L/mol K x 298 K

π = 3.74 atm

Answer: π = 3.74 atm

4 0

1 year ago

Choose the description that matches the law of included fragments.

GREYUIT [131]

Answer:

The answer is B

Explanation:

3 0

2 years ago