Ask question

Ask question

All categories

2 years ago

9

A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of n

e is added?

2 answers:

V125BC [204]2 years ago

8 0

From the explanation, we are given that:

The mole of gas n1 = 0.65 mol
The volume it occupied V1 = 900.0 ml
Mole of Neon = 0.35 ml

<h2>Further Explanation</h2>

To solve this question, we have to consider the Avogadro’s law, which state that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules”

Therefore, we have to calculate the total number of moles of the gases.

$(0.65 + 0.35) mol$

$1.00 mol$

Also we know that the volume it occupied V2 = unknown

According to the Avogadro’s law, the equation will be

$0.65 mol$ x $900.0ml$ = $1.00ml$ x $V2$

$585ml$ = $1.00 ml$ x $v2$

$V2$ = $585ml$

To convert 585ml to litre, then we have to divide the derived value by 1000

Therefore we have

$\frac{ 585ml}{1000}$

0.6L

Therefore, the volume of gas in liters that syringe will hold if 0.35 moles of neon is added is 0.6L

LEARN MORE:

A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of ne is added? brainly.com/question/6860167
A syringe contains 0.65 moles of he gas that occupy 750.0 ml. brainly.com/question/6442444

KEYWORDS:

moles
syringe
convert
litres
avogradro's law

Grace [21]2 years ago

5 0

<span>PV=nRT= a universal constant For any condition P1V1/n1T1=R and P2V2/n2T2=R i.e P1V1/n1T1=P2V2/n2T2 Becomes V1/n1=V2/n2 rearranging and solving V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>

You might be interested in

Which statements are true of the electron cloud model? Check all that apply. It is also known as the planetary model. It is cons

barxatty [35]

Answer:

-It is considered the modern atomic model.

-It describes the probable locations of the electrons

Explanation:

edge 2020

8 0

2 years ago

Read 2 more answers

Which of the following does not involve colligative properties?

fomenos

Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc.
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.

7 0

2 years ago

Read 2 more answers

You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu

ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L

4 0

1 year ago

A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa

bogdanovich [222]

Explanation:

The given data is as follows.

$V_{1}$ = 85.0 ml, $M_{1}$ = 0.9 M

$V_{2}$ = 85.0 ml, $M_{1}$ = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So, No. of moles = Molarity × Volume

= $0.9 M \times 0.085 L$ (as 1 L = 1000 ml so, 85 ml = 0.085 L)

= 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

$56.2 kJ/mol \times 0.076 mol$

= 4.271 kJ

or, = 4271 J (as 1 kJ = 1000 J)

Therefore, heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence, mass of HCl = density × Volume of HCl

= 1.00 g/ml × 85.0 ml

= 85 g

Similarly, mass of KOH = = density × Volume of HCl

= 1.00 g/ml × 85.0 ml

= 85 g

Hence, total mass of the solution = 85 g + 85 g

= 170 g

Also, q = $mC \Delta T$

4271 J = $170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C$

0.0773 = $T_{f} - 18.24$

$T_{f}$ = $18.317^{o}C$

Thus, we can conclude that final temperature of the mixed solution is $18.317^{o}C$ .

6 0

2 years ago

What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

5 0

2 years ago