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Mariana [72]
3 years ago
12

what was the initial pressure of 14.8 mL of gas at 75.5 C , when it occupied a volume of 16.5 mL at 70.2 C and 101.3 kPa?

Chemistry
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

1038.96 kPa

Explanation:

We’ll use the ideal gas law; P1V1/T1 = P2V2/T2

P1*14.8/75.5 = 101.3*16.5/70.2

P1 = (101.3 * 16.5 * 75.5) / (70.2 *14.8)

P1 = 1038.96

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a graduated cylinder (approximate as a regular cylinder) has a radius of 1.045 cm and a height of 30.48 cm. what is the volume o
Luba_88 [7]

Answer: 103.5 cm^3

Explanation:

8 0
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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


5 0
2 years ago
Is it called when a substance enters a gaseous phase without boiling?
Helen [10]

Answer:

its called sublimation

3 0
3 years ago
Read 2 more answers
Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured
yan [13]

Answer:

Water will boil at 76^{0}\textrm{C}.

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

                                         ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]

where, P_{1} and P_{2} are vapor pressures of liquid at T_{1} (in kelvin) and T_{2} (in kelvin) temperatures respectively.

Here, P_{1} = 760.0 mm Hg, T_{1} = 373 K, P_{2} = 314.0 mm Hg

Plug-in all the given values in the above equation:

                          ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]

                       or, T_{2}=349 K

So, T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}

Hence, at base camp, water will boil at 76^{0}\textrm{C}

6 0
3 years ago
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