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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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If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting soluti
GaryK [48]

Answer : The boiling point of the resulting solution is, 100.6^oC

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

or,

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^oC

k_b = boiling point constant  = 0.52^oC/m

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute (sucrose) = 5.0 g

w_1 = mass of solvent (water) = 10.0 g

M_2 = molar mass of solute (sucrose) = 342.3 g/mol

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

(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}

T_b=100.6^oC

Therefore, the boiling point of the resulting solution is, 100.6^oC

5 0
3 years ago
Even though the para position is one carbon farther from the carboxy group than the meta position, p-cyanobenzoic acid is more a
Nuetrik [128]

Answer:

Dissociation of O-H bond is highly favorable in p-cyanobenzoic acid resulting to higher acidity as compared to m-cyanobenzoic acid.

Explanation:

In p-cyanobenzoic acid, adjacent carbon to carboxyl group (-COOH) gets a partial positive charge due to electron withdrawing resonating effect of carboxyl group as compared to m-cyanobenzoic acid.

Due to this partial positive charge on carbon atom in p-cyanobenzoic acid, O-H bond in -COOH group remains highly polarized towards oxygen atom. Hence, dissociation of O-H bond is highly favorable in p-cyanobenzoic acid resulting to higher acidity as compared to m-cyanobenzoic acid.

Resonance structures are given below.

6 0
3 years ago
How does the type of liquid used to make ice cubes affect the rate at which it melts?
3241004551 [841]
It doesn’t affect it
5 0
3 years ago
The element of 2.345​
JulijaS [17]

Answer:

Yttrium

Explanation:

7 0
3 years ago
A buffer made with 100.00 mL of 0.95 M lactic acid (Ka=1.4x10-4) and 200.00 mL of 0.50 M lactate has a final volume of 1.0 L. Wh
Art [367]

Answer:

3.77

Explanation:

The pH of a buffer can be calculated by Handerson-Halsebach equation:

pH = pKa+ log [A⁻]/[HA]

Where pKa = -logKa, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In this case [A⁻] = concentration of lactate. Because the final volume will be the same, and [X] = mol/L, we can use the number of moles instead of the concentration.

nHA = 0.1 L * 0.95 mol/L = 0.095 mol

nA⁻ = 0.2 L * 0.5 mol/L = 0.1 mol

When HCl is added, it will dissociate in H⁺ and Cl⁻. H⁺ will react if A⁻ to form more HA, so the equilibrium will be shift. Because of that, the number of moles of HA will be the initial plus the number of moles of H⁺ added (which is equal to the number of moles of HCl), and the number of moles of A⁻ will be the initial less the number of moles of H⁺.

nH⁺ = nHCl = 0.015 L* 0.75 = 0.01125 mol

nA⁻ = 0.1 - 0.01125 = 0.08875 mol

nHA = 0.095 + 0.01125 = 0.10625 mol

pKa = -log(1.4*10⁻⁴) = 3.85

pH = 3.85 + log(0.08875/0.10625)

pH = 3.77

8 0
3 years ago
Read 2 more answers
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