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dimulka [17.4K]
4 years ago
13

What are examples of substances with a high viscosity

Chemistry
2 answers:
Zarrin [17]4 years ago
5 0

Honey, syrup, glue can i be marked brainliest  

ipn [44]4 years ago
4 0

some examples for this question are: Honey Or water.

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over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th
Aleonysh [2.5K]

Answer : The moles of methane gas could be, 7.90\times 10^{-3}mol

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}

[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}

where,

R_1 = rate of effusion of fluorine gas

R_2 = rate of effusion of methane gas

n_1 = moles of fluorine gas = 5.13\times 10^{-3}mol

n_2 = moles of methane gas = ?

t_1=t_2 = time = 12.3 min  (as per question)

M_1 = molar mass of fluorine gas  = 38 g/mole

M_2 = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}

n_2=7.90\times 10^{-3}mol

Therefore, the moles of methane gas could be, 7.90\times 10^{-3}mol

8 0
3 years ago
Which is the correct way to write 602,200,000,000,000,000,000,000 in scientific notation?
sladkih [1.3K]

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Balance the redox reaction equation (occurring in acidic solution) and choose the correct coefficients for each reactant and pro
algol13
Balanced chemical reaction: 
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Oxidation half-reaction: Sn </span>→ Sn²⁺ + 2e⁻.<span>
Reduction half-reaction: PbO</span>₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O.
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Oxidation is increase of oxidation number, reduction is decrease of oxidation number.
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