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enyata [817]
3 years ago
5

Once a slide is prepared and placed onto the microscope, the magnification and focus need to be altered. Describe how to change

the magnification and bring the image into focus using a light microscope like the one in the simulation.
Chemistry
2 answers:
Harlamova29_29 [7]3 years ago
5 0

Answer:

我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈

Explanation:

我實際上不知道答案,我只是為了點我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈數而這樣做,哈哈,祝你好運哈哈

Evgen [1.6K]3 years ago
5 0

Answer:

They said: I don’t actually know the answer, I just do it for points, haha, good luck haha

Explanation:

I actually don’t know the answer, I’m just for the point I don’t actually know the answer, I just do it for the point, haha, good luck haha ​​and do this, haha, good luck haha

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7 0
3 years ago
Which mountain range was once the tallest in the world?
Ne4ueva [31]

Answer:

A

Explanation:

7 0
4 years ago
A sample of water is mixed with a surfactant. What will most likely happen to the viscosity of the water?
sp2606 [1]
If a sample o water is mixed with a surfactant, the most likely result is that the viscosity of the water will increase. This is because of the property of the surfactant which is generally viscous; and becomes more viscous when mixed with water. 
5 0
4 years ago
A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature
Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

  • v_0=500
  • \alpha =1.12\times 10^{-4}
  • \Delta \epsilon = 25

Now,

Increase in volume will be:

⇒ \Delta V = \alpha\times v_0\times \Delta \epsilon

By putting the given values, we get

           =1.12\times 10^{-4}\times 500\times 25

           =1.12\times 10^{-4}\times 12500

           =1.4  \ mL

8 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
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