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3 years ago

What state of matter is at 35 Celsius

1 answer:

7 0

Different forms of matter have different melting/boiling points. For example, at 100 degrees Celsius, H2O (water) will turn from lliquid to gas. But NaOH (table salt) doesn't even go from solid to liquid until some 800 degrees Celsius. So, in order to figure out which state matter is at 35 Celsius, you'd have to be more specific about what kind of matter...

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi

Gelneren [198K]

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

3 0

3 years ago

A gas exerts a pressure of 0.62 atm. Convert this to kPa and mmHg. Be sure to show your work.

IRINA_888 [86]

Answer:

The answer to your question is 0.62 atm = 62.82 kPa = 471.2 mmHg

Explanation:

Data

P = 0.62 atm

P = ? kPa

P = ? mmHg

Process

1.- Look for the conversion factor of atm to kPa and mmHg

1 atm = 101.325 kPa

1 atm = 760 mmHg

2.- Do the conversions

1 atm ----------------- 101.325 kPa

0.62 atm ------------ x

x = (0,62 x 101.325) / 1

x = 62.82 kPa

1 atm ------------------ 760 mmHg

0.62 atm ------------ x

x = (0.62 x 760)/1

x = 471.2 mmHg

4 0

3 years ago

what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3

ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

$Q=mc\Delta T$

$\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}$

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

3 0

3 years ago

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kobusy [5.1K]

Answer is c...............

6 0

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What are the functions of protein in organisms?

Juliette [100K]

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3 years ago