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White raven [17]
3 years ago
9

If a tire can hold 19 L of air at 175 kPa and 25° Celsius, what volume of air can the tire hold at 215 kPa and 30° Celsius?

Chemistry
1 answer:
Usimov [2.4K]3 years ago
6 0
Using p1v1/t1=p2v2/t2
p1=175kpa
p2=215kpa
v1=19l
v2=?
t1=25+273=298
t2=30+273=303
175x19/298=215xv2/303
8.808x303=215v2
2669/215=v2
v2=12.4l
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frutty [35]
Water has a chemical formula of H2O. This means that for every 2 moles of hydrogen and 1 mole of oxygen, one mole of water will be formed.

Note that hydrogen gas and oxygen gas are both biatomic molecules. 

 (1)                   (182 mol H2) x (1 mol H2O/ 1 mol H2) = 182 mol H2O
 (2)                   (86 mol O2) x (2 mol H2O / 1 mol O2) = 172 mol H2O

We choose the smaller number of the two as the answer to this item. Thus, the answer to this question is 172 mol of H2O can be formed out of the given quantities. 
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3 years ago
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Answer:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

Explanation:

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The superscript of \rm Ag^{1+} indicates that each of these ions carries a charge of +1. That corresponds to the shortage of one electron for each \rm Ag^{+} ion.

Similarly, the superscript +3 on each \rm Al^{3+} ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of \rm Ag^{+} (among the reactants) is x, and that the coefficient of \rm Al^{3+} (among the reactants) is y.

\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag.

There would thus be x silver (\rm Ag) atoms and y aluminum (\rm Al) atoms on either side of the equation. Hence, the coefficient for \rm Al\! and \rm Ag\! would be y\! and x\!, respectively.

\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag.

The x \rm Ag^{1+} ions on the left-hand side of the equation would correspond to the shortage of x electrons. On the other hand, the y Al^{3+} ions on the right-hand side of this equation would correspond to the shortage of 3\, y electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of x electrons, the right-hand side should also be x\! electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of 3\, y electrons. These two expressions should have the same value. Therefore, x = 3\, y.

The smallest integer x and y that could satisfy this relation are x = 3 and y = 1. The equation becomes:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

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A compound is found to contain 7.962 % silicon , 20.10 % chlorine , and 71.94 % iodine by mass. to answer the question, enter th
arlik [135]
Empirical   formula  is  calculated   as  follows

calculate  the  moles  of  each  element,  that  is   %  composition/  molar  mass
molar  masses  (  Si=  28.09g/mol ,  Cl=  35.5  g/mol,  I=126.9 g/mol)

moles  of  silicon =  7.962/28.09g/mol=  0.283  moles
moles  of  chlorine =  20.10 / 35.5g/mol =  0.566  moles
moles  of  iodine=  71.94 / 126.9  g/mol=  0.567  moles

divide  each  mole   with  smallest    mole  (0.283)
that  is    silicon =  0.283/0.283= 1 mole
             chlorine =  0.566/0.283=  2 mole
            Iodine=  o.567/0.283= 2  moles
empirical  formula  is  therefore=  SiCl2I2


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