As the energy of the molecules increases, the motion of the molecules ______

Commercial manufacturers produce nitric acid (HNO3) by the Ostwald process. The process requires three steps:

Shkiper50 [21]

Answer:

this doesnt make sence ezxplain the subject

Explanation:

5 0

3 years ago

Significant Figures: How many significant figures<br> are in 865,010?

Harrizon [31]

Answer: There are five significant figures in 865,010.

Explanation:

When a degree of accuracy is stated by each digit present in a mathematical figure then it is called a significant figure.

Rules for counting significant figures is as follows.

Any non-zero digits and zeros present between a non-zero figure are counted. For example, 3580009 has seven significant figures.

Trailing zeros are counted in a non-zero figure. For example, 0.00250 has three significant figures.

Leading zeros are not counted. For example, 0.0025 has two significant figures.

So, in the given figure 865010 has five significant figures and the trailing zero will not be counted.

Thus, we can conclude that there are five significant figures in 865,010.

7 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$