12. Gavin is having a difficult time

Chemistry

1 answer:

schepotkina [342]3 years ago

5 0

Answer:

Friction

Explanation:

the resistance that one surface or object encounters when moving over another.

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Any time an atom transfers or shares an electron a is formed

polet [3.4K]

Anytime an atom shares or transfers electrons a bond is formed, sharing= covalent and transfer= ionic

6 0

3 years ago

Ethanol (c2h5oh) melts at -114Â°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9

Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid $C_2H_5OH=-135^oC=138 K(0^oC=273K)$

Melting temperature of Solid $C_2H_5OH=114^oC=159 K$

Temperature of liquid $C_2H_5OH=-50^oC=223K$

Specific heats of solid ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid $C_2H_5OH$ at 159 K

$\Delta T_1$ = 159 K - 138 K = 21 K

$Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J$

Heat required to melt and raise the temperature of $C_2H_5OH$ upto 223 K

$\Delta T_2$ = 223 K - 159 K = 64 K

$Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J$

Total heat to convert solid ethanol to liquid ethanol at given temperature :

$Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ$ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0

3 years ago

There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore

devlian [24]

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

$\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\$