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RoseWind [281]
2 years ago
15

Compare and contrast the Andes Mountains and the Rocky Mountains. How are they formed? What consumers and producers are in each?

.
Chemistry
1 answer:
Naily [24]2 years ago
7 0

Answer:

ummm i need help too lol

Explanation:

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Evaluate three communication skills that can assist you in sustaining positive relationships​
Tpy6a [65]

1. Accept and celebrate differences. One of the biggest challenges we experience in relationships is that we are all different. We can perceive the world in many ways. Certainly astumbling block that we come across when we try to build relationships is a desire or an expectation that people will think like we do and, in this way, it is so much easier to create a rapport. We feel more comfortable when we feel that people “get” us and can see our point of view. Life, however, would be very dull if we were all the same and, while we may find it initially easier, the novelty of sameness soon would wear off. So accepting and celebrating that we are all different is a great starting point.

2. Listen effectively. Listening is a crucial skill in boosting another person’s self-esteem, the silent form of flattery that makes people feel supported and valued. Listening and understanding what others communicate to us is the most important part of successful interaction and vice versa.

Active or reflective listening is the single most useful and important listening skill. In active listening, we also are genuinely interested in understanding what the other person is thinking, feeling, wanting, or what the message means, and we are active in checking out our understanding before we respond with our own new message. We restate or paraphrase our understanding of their message and reflect it back to the sender for verification. This verification or feedback process is what distinguishes active listening and makes it effective.

3. Give people your time. Giving time to people is also a huge gift. In a world where time is of the essence and we are trying to fit in more than one lifetime, we don’t always have the time to give to our loved ones, friends, and work colleagues. Technology has somewhat eroded our ability to build real rapport and we attempt to multi-task by texting and talking at the same time.

Being present in the time you give to people is also important, so that, when you are with someone, you are truly with someone and not dwelling in the past or worrying about the future. The connection we make with other people is the verytouchstone of our existence, and devoting time, energy, and effort to developing and building relationships is one of the most valuable life skills.

3 0
3 years ago
A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol
Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

\Delta T = change in temperature = T_2-T_1=(26.3-20.2)^oC=6.1^oC

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

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

q=2623.61J

Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = Concentration\times Volume=1M\times 0.050L=0.050mole

\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0
3 years ago
What are some applications of quantities analysis?
andriy [413]

Answer:

Needless to say, we have faced a lot of challenges in the analysis and study of such a huge volume of data with the traditional data processing tools. To overcome these challenges, some big data solutions were introduced such as Hadoop. These big data tools really helped realize the applications of big data.

Explanation:

6 0
3 years ago
In the laboratory you dissolve 19.1 g of ammonium fluoride in a volumetric flask and add water to a total volume of 375 . mL. Wh
11111nata11111 [884]

Answer:

1.376 M

Explanation:

The following data were obtained from the question:

Mass of ammonium fluoride (NH₄F) = 19.1 g

Volume of solution = 375 mL

Molarity of ammonium fluoride (NH₄F) =?

Next, we shall convert 375 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

375 mL = 375 mL × 1 L / 1000 mL

375 mL = 0.375 L

Next, we shall determine the number of mole in 19.1 g of ammonium fluoride (NH₄F). This can be obtained as follow:

Mass of NH₄F = 19.1 g

Molar mass of NH₄F = 14 + (4×1) + 19

= 14 + 4 + 19

Molar mass of NH₄F = 37 g/mol

Mole of NH₄F =?

Mole = mass /Molar mass

Mole of NH₄F = 19.1 / 37

Mole of NH₄F = 0.516 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Volume of solution = 0.375 L

Mole of NH₄F = 0.516 mole

Molarity of NH₄F =?

Molarity = mole /Volume

Molarity of NH₄F = 0.516 / 0.375

Molarity of NH₄F = 1.376 M

Therefore, the molarity of the ammonium fluoride (NH₄F) is 1.376 M

3 0
2 years ago
mixture of 75 cm3of oxygen and 12.5 cm3 of a gaseous hydrocarbon H were exploded in an eudiometer. After cooling to room tempera
Mrac [35]

Answer:

molar mass of C₃H ₈ = 44 g/mole

Explanation:

Computation of the amount of oxygen that reacts .

⇒ 75 - 12.5

⇒ 62.5 cm³

Computation of  proportion of hydrocarbons

2 mole hydrocarbons 3n+1 oxygen

⇒ 3 n + 1 = [62.5 × 2] / 12.5

⇒ 3 n + 1 = 10

⇒ n = 3

So,

Formula of the hydrocarbon is C₃H ₈

Computation of molar mass of C₃H ₈

⇒ [12×3] + [1×8]

⇒ 44 g/mole

8 0
3 years ago
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