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3 years ago

Volume occupied 3.52x10^32 moluchlesof Mathane (CH4)1) At STP

Chemistry

1 answer:

Naddika [18.5K]3 years ago

6 0

Answer:

volume = 13097674418.528dm³

Explanation:

n = (3.52)*10^32/(6.02)*10^23)

n = (584717607.97)

n = volume /molar volume

molar volume at stp = 22.4dm³

volume= 584717607.97 x 22.4

volume = 13097674418.528dm³

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What is the ratio (in the form of a decimal) of the root-mean-square speed of 238UF6 to that of 235UF6 at constant temperature?

NNADVOKAT [17]

Answer:

ratio=0.996m/s

Explanation:

$RMS=\sqrt{\frac{3RT}{M} }$ ; M= molecular weight of compound;

RMS speed is inversly proportional to the molecular weight hence compound having less molecular weight will have more rms speed value.

$V1 for 238 UF_{6}; M1=238+114=352g/mole$

$V1 for 235 UF_{6}; M1=238+114=352g/mole$

$V1=\sqrt{\frac{3RT}{352} }$ ;

$V2=\sqrt{\frac{3RT}{349} }$ ;

$\frac{V1}{V2}$ =ratio of $238UF_{6} F6 to 235UF_{6}$ = $\sqrt{\frac{349}{352} }$ ;

$\frac{V1}{V2} =0.9957m/s;$

$\frac{V1}{V2} =0.996m/s;$

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3 years ago

Calculate the formula/molecular mass of following

ddd [48]

Answer:

Refer to the period table. Use the atomic mass for calculation. Multiply the atomic mass with the number of atoms. For example, Al2(SO4)3. (27 × 2) + (32 x 3) + (16 x 12) = molecular mass

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3 years ago

An article in Science News stated, "Noble gases are snobs." What did the author mean?

eduard

Answer:

Explanation:

Noble gases occupy the last group of the periodic table

They have fully filled valence electron shell

What this means is that they have attained stability and thus do not take part in chemical bonding that usually invloves the transfer or sharing of electrons

Thus, noble gases are snobs because they do not partake in chemical bonding with atoms of other elements or atoms of theirselves

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1 year ago

3. A student adds 0.400g of n-propanol to 13.0 g of t-butanol. What is the molality of the solution? Show your calculations. (3

marissa [1.9K]

Answer:

$m=0.512m$

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

$n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol$