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Elza [17]
3 years ago
7

All of the following describe the unique role of water except-

Chemistry
1 answer:
9966 [12]3 years ago
6 0
Option c
water is non polar and can dissolve non polar compounds such as oil
water is unable to dissolve or mix with substances like oil therefore making the statement false
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What is the empirical formula for ribose?
BartSMP [9]

Answer:

The empirical formula of ribose (a sugar) is CH2O. In a separate experiment, using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol.Explanation:

5 0
3 years ago
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How many cups of beer are in a keg?
Alinara [238K]
The answer is roughly 165 cups of beer.
5 0
3 years ago
How many particles would be found in a 1.224 g sample of K2O
lbvjy [14]

Answer:

9.96*10^21

Explanation:

Molar mass of K2O=29*2+16

= 74g per mol

number of moles in the sample= 1.224/ 74

=0.1654

Number of particles in 1 mole=6.0221409*10^23

Number of particles= 0.01654*6.0221409*10^23

=9.96*10^21

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3 years ago
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Which model of the atom was used as a result of JJ Thomson's cathode ray
mariarad [96]

Answer:

a. Plum pudding model

Explanation:

The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.

J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.

  • This led him to suggest the plum pudding model of the atom.
  • The model reflects electrons being surrounded by a volume of negative charges.
6 0
3 years ago
If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

4 0
2 years ago
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