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pashok25 [27]
3 years ago
8

1. Which of the following is a linear inequality in two variables?

Chemistry
2 answers:
dimulka [17.4K]3 years ago
8 0

Answer:

I hope I can help you

Explanation:

I hope I can help you

Stolb23 [73]3 years ago
3 0

Answer:

  1. A. 7x - 4y = 3
  2. C. 6p 2
  3. C. 2
  4. A. (4, 5)
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What is the Dependent variable?
horsena [70]

Explanation:

Just like an independent variable, a dependent variable is exactly what it sounds like. It is something that depends on other factors. For example, a test score could be a dependent variable because it could change depending on several factors such as how much you studied, how much sleep you got the night before you took the test, or even how hungry you were when you took it. Usually when you are looking for a relationship between two things you are trying to find out what makes the dependent variable change the way it does.

7 0
3 years ago
For the reaction shown, calculate how many moles of NH3 form when 16.72 moles of reactant completely reacts:
Agata [3.3K]

Answer : The moles of NH_3 formed are, 22.3 moles.

Explanation : Given,

Moles of N_2H_4 = 16.72 mol

The given chemical reaction is:

3N_2H_4(l)\rightarrow 4NH_3(g)+2N_2(g)

From the balanced chemical reaction, we conclude that:

As, 3 moles of N_2H_4 react to give 4 moles of NH_3

So, 16.72 moles of N_2H_4 react to give \frac{4}{3}\times 16.72=22.3 moles of NH_3

Therefore, the moles of NH_3 formed are, 22.3 moles.

8 0
3 years ago
A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re
sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula  -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass  = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature  10.0°C + 273 = 283 K

T₁ = initial temperature ,  40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ )  = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0
3 years ago
I really need your help best answer gets the brainiest! please answer its 50 POINTS!
dangina [55]

Answer:

add it all up

Explanation:

6 0
3 years ago
The group in an experiment that is not exposed to the tested variable is called the group
kati45 [8]

Answer:

yes your answer is correct for this question.

5 0
3 years ago
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