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tester [92]
3 years ago
14

Consider two gases—oxygen and nitrogen at STP. How will the kinetic energy of the gases be affected if we only increase the temp

erature of oxygen?
The kinetic energy of both gases will increase, and then after sometime both will have the same kinetic energy.
Oxygen will have more kinetic energy than nitrogen.
Oxygen will have less kinetic energy than nitrogen.
Initially nitrogen will have more kinetic energy, and then after sometime oxygen will have more kinetic energy.
Kinetic energy won’t be affected at all.
Chemistry
2 answers:
rewona [7]3 years ago
6 0
(B) Oxygen will have more kinetic energy than nitrogen.
alex41 [277]3 years ago
3 0

The correct option is this: OXYGEN WILL HAVE MORE KINETIC ENERGY THAN NITROGEN.

Increasing the temperature of oxygen requires the application of heat. The heat energy that is applied to the gas will make the particles of the oxygen gas to gain more kinetic energy and to move more rapidly than before, by so doing, the particles will colloid more with one another and with the wall of the container. The kinetic energy of the particles of the nitrogen gas will remain the same since its temperature was not affected.

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Aluminum metal reacts with iron (III) oxide to produce aluminum oxide
Ira Lisetskai [31]

Answer:

is this a T or F question? if so then T

Explanation:

7 0
3 years ago
Please help me solve this!
yulyashka [42]

Answer : The image is attached below.

Explanation :

For O_3:

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = \frac{m}{M}=\frac{24g}{48g/mol}=0.5mol

Number of particles, N = n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}

For NH_3:

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = \frac{m}{M}=\frac{170g}{17g/mol}=10mol

Number of particles, N = n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}

For F_2:

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = \frac{m}{M}=\frac{38g}{38g/mol}=1mol

Number of particles, N = n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}

For CO_2:

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = n\times M=0.10mol\times 44g/mol=4.4g

Number of particles, N = n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}

For NO_2:

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = n\times M=0.20mol\times 46g/mol=9.2g

Number of particles, N = n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}

For Ne:

Molar mass, M = 20 g/mol

Number of particles = 1.5\times 10^{23}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol

Mass, m = n\times M=0.25mol\times 20g/mol=5g

For N_2O:

Molar mass, M = 44 g/mol

Number of particles = 1.2\times 10^{24}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol

Mass, m = n\times M=1.9mol\times 44g/mol=83.6g

For unknown substance:

Number of particles = 3.0\times 10^{23}

Mass, m = 8.5 g

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol

Molar mass, M = \frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol

The substance is NH_3.

3 0
3 years ago
Describe the trend (pattern) in the numbers of mobile phone owners in the Uk since 1990.
Inessa05 [86]

Explanation:

<em><u>SMARTPHONE OWNERSHIP IS GROWING RAPIDLY AROUND THE WORLD, BUT NOT ALWAYS THE SAME.</u></em>

In emerging economies, the use of technology is still more common among the young and the educated

A farmer takes a selfie with a smartphone at a rally in Jaipur, India. (Vishal Bhatnagar/AFP/Getty Images)

A farmer takes a selfie with a smartphone at a rally in Jaipur, India. (Vishal Bhatnagar/AFP/Getty Images)

The chart shows that smartphone ownership in advanced economies is higher than in emerging economies.

Mobile technology has spread rapidly around the world. Today, it is estimated that more than 5 billion people have mobile devices, and more than half of these connections are smartphones. But growth in mobile technology so far has been unequal, either nationwide or within it. People in advanced economies are more likely to have mobile phones - smartphones in particular - and more likely to use the internet and social media than people in emerging economies. For example, a median of 76% across the 18 advanced economies surveyed have a smartphone, compared to a median of only 45% in emerging economies.

Smartphone ownership can vary by country, and even across developed economies. While about nine in ten or more South Koreans, Israelis and the Netherlands own a smartphone, the ownership rate is closer to six in ten in other developed countries such as Poland, Russia and Greece. In emerging economies as well, smartphone ownership rates vary significantly, from a high of 60% in South Africa and Brazil to just around four in ten in Indonesia, Kenya and Nigeria. Among the countries surveyed, ownership was lowest in India, where only 24% reported having a smartphone.

4 0
2 years ago
Help i really don’t understand
sattari [20]
1. 1mL
2. 2mL
3. 0.2mL
4. 0.5mL
7 0
3 years ago
3.) 2.46 grams of Cu(NO3)2 would contain how many formula units?
RUDIKE [14]

Answer:

7.90×10²¹ formula units

Explanation:

From the question given above, the following data were obtained:

Mass of Cu(NO₃)₂ = 2.46 g

Formula units of Cu(NO₃)₂ =?

From Avogadro's hypothesis,

1 mole of Cu(NO₃)₂ = 6.02×10²³ formula units

Next, we shall determine the mass of 1 mole of Cu(NO₃)₂. This can be obtained as follow:

1 mole of Cu(NO₃)₂ = 63.5 + 2[14 + (3×16)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g

Thus,

187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units

Finally, we shall determine the formula units contained in 2.46 g of Cu(NO₃)₂. This can be obtained as follow:

187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units.

Therefore,

2.46 g of Cu(NO₃)₂ =

(2.46 × 6.02×10²³)/187.5

= 7.90×10²¹ formula units

Thus, 2.46 g of Cu(NO₃)₂ contains 7.90×10²¹ formula units

7 0
3 years ago
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