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lutik1710 [3]
3 years ago
6

A 0.55 g sample of H20 contains how manymolecules of water?Answer in units of molec.​

Chemistry
1 answer:
noname [10]3 years ago
7 0

Answer:

Explanation:

Num of molecules = num of moles * Avogadro's constant (6.02* 10^23)

But num of moles = reacting mass / molar mass

Molar mass of H20= 2*1 + 16 = 2+16 = 18g

Reacting mass of H20 = 0.55g

Therefore, num of moles of H20 = 0.55g/18g = 0.031 moles

Therefore, num of molecules of H20 = 0.031 * 6.02*10^23

= 1.87*10^22 molecules of H20

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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Let's assume the gas behaves ideally.

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                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

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3 years ago
Iodine-131 half life 8.040 how long will it take for a 40 gram sample of iodine-131 to decay to 0.75 grams
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Answer:

See explanation below

Explanation:

To get this, we need to apply the general expression for half life decay:

N = N₀e(-λt)    (1)

Where:

N and N₀ would be the final and innitial quantities, in this case, masses.

t: time required to decay

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From the above expression we need λ and t. To get λ we use the following expression:

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And we have the value of half life, so, replacing we have:

λ = 8.04 / ln2 = 11.6

Now, we can replace in (1) and then, solve for t:

0.75 = 40 exp(-11.6t)

0.75 / 40 = exp(-11.6t)

ln(0.01875) = -11.6t

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t = -3.9766 / -11.6

<h2>t = 0.34 days</h2><h2></h2>
3 0
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