All categories

3 years ago

Question is in picutree eee eee e e

Chemistry

2 answers:

gtnhenbr [62]3 years ago

7 0

C flip the switch up

Serga [27]3 years ago

3 0

Answer:

its is a or c I'm more confident in a though

You might be interested in

In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696 = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 4.81 moles

4 0

3 years ago

Which statement accurately describes the illustration?

jasenka [17]

I'm not sure but I think it is A. image one has the most spread out particles like a gas, and b has closer together particles like a liquid or solid. since there are no choices that say A=gass and B=solid, so I am guessing it is answer A.

4 0

3 years ago

Read 2 more answers

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

Which sample of matter is classified as a solution

svet-max [94.6K]

Answer: Air, sea water, and carbonation dissolved in soda are all examples of homogeneous mixtures, or solutions. Hope this helps :)

7 0

3 years ago

If 17.4 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentra

Verizon [17]

NH3 +HCl ----> NH4Cl

moles of HCl used = (0.8 x 17.4) /1000= 0.0139 moles
by use of reacting ratio between HCl to NH4Cl which is 1:1 therefore the moles of NH4Cl is also = 0.0139 moles
molar concentration = moles /volume in liters

molar concentration is therefore= (0.0139/5) x1000 = 2.7 M

8 0

3 years ago

Read 2 more answers