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3 years ago

Why are cirrocumulus clouds sometimes called mackerel clouds

Chemistry

1 answer:

evablogger [386]3 years ago

4 0

Cirrocumulus clouds look similar to fish scales it just pops over different kind of shapes. I hope this helps(:

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Mixtures can be separated by physical changes. Please select the best answer from the choices provided

Mazyrski [523]

Where are the choices? :/

4 0

3 years ago

Read 2 more answers

How many cobalt atoms are in 345 g of cobalt

lilavasa [31]

Answer:

$\boxed {\boxed {\sf About \ 3.53 \ *10^{24}atoms \ Co}}$

Explanation:

To convert from grams to atoms:

Convert grams to moles
Convert moles to grams

1. Convert grams to moles

First, find the molar mass of cobalt using the Periodic Table of Elements.

Cobalt (Co): 58.93319 g/mol

Next, use this mass as a ratio or fraction.

$\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Multiply the mass of the given sample (345 grams) by this ratio.

$345 \ g \ Co *\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Flip the fraction so the grams of Cobalt will cancel each other out when multiplying.

$345 \ g \ Co *\frac{1 \ mol \ Co}{58.93319 \ g\ Co}$

$345 \ *\frac{1 \ mol \ Co}{58.93319 }$

$\frac{345 \ mol \ Co}{58.93319 } =5.854086636 \ mol \ Co$

2. Convert moles to atoms

Use Avogadro's Number, which tells us the number of units (in this case atoms) in 1 mole.

$6.022 *10^{23} \ atoms/mol$

Use this number as a ratio or fraction.

$\frac{6.022 * 10^{23} \ atoms \ Co}{1 \ mol \ Co}}$

Multiply this ratio by the number of moles we found.

$5.854086636 \ mol \ Co*\frac{6.022 * 10^{23 }\ atoms \ Co}{1 \ mol \ Co}}$

The moles of Cobalt will cancel.

$5.854086636 \ *\frac{6.022 * 10^{23 }\ atoms \ Co}{1 }}$

$5.854086636 \ *{6.022 * 10^{23 }\ atoms \ Co}$

$3.52533097*10^{24} \ atoms \ Co$

3.Round

The original measurement of 345 has 3 significant figures (3, 4, and 5). We must round to 3 sig figs, which is the hundredth place for this measurement.

$3.52533097*10^{24} \ atoms \ Co$

The 5 in the thousandth place tells us to round the 2 to a 3.

$3.53 \ *10^{24}atoms \ Co$

There are about 3.53*10²⁴ atoms of cobalt in 345 grams.

6 0

3 years ago

Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 ∘C, the vapor pressure of pure be

Kisachek [45]

Answer:

$P_{solution} = 85.3Torr$

Explanation:

Raoult's law is a tool that allows to determine vapour pressure of solutions. The formula is:

$P_{solution} = X_{solvent}P_{solvent}$ (1)

Where

P is Pressure of solution and solvent and X is mole fraction.

Moles of solute and solvent are:

Biphenyl:

11.5g×(1mol /154.21g) = 0.0746mol

Benzene

31.9g×(1mol /78.11g) = 0.408mol

Mole fraction of benzene is:

$\frac{0.408mol}{0.408mol + 0.0746mol}$ = 0.846

Replacing in (1):

$P_{solution} = 0.846*100.84Torr$

$P_{solution} = 85.3Torr$

I hope it helps!

3 0

3 years ago

Describe the location of the Philippines

allsm [11]

Answer:

Philippines, island country of Southeast Asia in the western Pacific Ocean. It is an archipelago consisting of more than 7,000 islands and islets lying about 500 miles (800 km) off the coast of Vietnam. Manila is the capital, but nearby Quezon City is the country's most-populous city.

5 0

3 years ago

Given that delta hvap is 58.2 kj/mol and the boiling point is 83.4 c 1atm if one mole of this substance is vaporized at 1atm cal

astraxan [27]

Change in Gibb's free energy of system (ΔG) = ΔH - TΔS.........(Eq. 1)
Now, if magnitude of ΔG <0, then reaction is spontaneous.
if magnitude of ΔG > 0, then reaction is non-spontaneous.
At equilibrium, ΔG = 0
When at boiling point, liquid state is in equilibrium with vapour state. Hence, it present case ΔG = 0

∴ Eq 1 becomes, ΔH = TΔS
here, ΔH = 58.2 kj/mol (Given),
∴ At T = 83.4 oC = 356.4 K, ΔS = 0.1633 kj/mol.K

3 0

3 years ago