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pogonyaev
2 years ago
8

Which of the following is an example of rapid combustion?

Chemistry
1 answer:
rjkz [21]2 years ago
6 0

phosphorus in water is rapid combustion.

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If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions
dimulka [17.4K]

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

P_2 = final pressure of dry gas at STP =  760 mm Hg

V_1 = initial volume of dry gas = 85.0 mL

V_2 = final volume of dry gas at STP = ?

T_1 = initial temperature of dry gas = 20^oC=273+20=293K

T_2 = final temperature of dry gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get the final volume of wet gas at STP

\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}

V_2=77.4mL

Volume of dry gas at STP is 77.4 mL.

5 0
3 years ago
The substances present before a chemical reaction takes place are called?
andrew11 [14]
The substances present before the reaction are the reactants.  (As the reaction goes through, the substances that are produced are called the products of the reaction).
5 0
3 years ago
Read 2 more answers
3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen​
Morgarella [4.7K]

Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

Thus, the total number of molecules turns out:

Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec

Regards.

6 0
3 years ago
Mosquitoes lays eggs that must hatch in water. The young, called larvae live their first few days in water
nevsk [136]

Answer:

Explanation:

I believe he is C

3 0
2 years ago
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th
oee [108]

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , n=\dfrac{2.666}{16}=0.167\ mole .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , m=12\times 0.167=2\ g .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

5 0
2 years ago
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