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Zigmanuir [339]
4 years ago
15

NEED THIS NOW PLZ!!!

Chemistry
2 answers:
Norma-Jean [14]4 years ago
7 0

Answer:

Centimeters per minute​

Explanation:

Speed measurement is formated as [distance] per [time], like miles per hour.

B and C are incorrect because they are [time] per [distance]

A is incorrect because it is [time] per [time]

vitfil [10]4 years ago
7 0

Answer : The correct option is (D) Centimeters per minute​

Explanation :

Speed : It is defined as the distance traveled by the object per unit time.

The unit of distance are, kilometer, meter, centimeter etc.

The unit of time are, minute, second, hour etc.

Formula used :

Speed=\frac{Distance}{Time}

From the given options, the correct measurement of speed is centimeters per minute​.

Hence, the correct option is (D) Centimeters per minute​

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0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

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Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
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The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
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Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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