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Ivanshal [37]
2 years ago
6

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

ature and pressure if you need a total of 1.36 mol of oxygen gas in the vessel? Express the volume to three significant figures and include the appropriate units.
Chemistry
1 answer:
fgiga [73]2 years ago
4 0

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

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Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

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2 years ago
Amy performed an experiment in lab. She improperly mixed the chemicals, and an explosion of light, sound, and heat occurred. Whe
agasfer [191]

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2 years ago
Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0
2 years ago
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Roman55 [17]
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What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (The values for Ka1 and Ka2 for ascorbic acid are 8.0×10−5 and
Tomtit [17]

Answer:

pH = 2.32

Explanation:

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Ka2 is very less so i am not considering that dissociation.

now Ka = 8.0×10−5

            = [H3O+] [HA-] / [H2A]

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solve the quardratic equation

X =0.004693 M

pH = -log[H+}

    = -log [0.004693]

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pH = 2.32

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