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notka56 [123]
3 years ago
13

What is the total magnification if we using objective lens 4x and our eyepiece (ocular) lens is 10x

Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
Answer: 40x
4x times 10x equals 40x
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What is a layer of solid of the solid earth<br>​
nikitadnepr [17]

The answer is The mantle

6 0
2 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
The mass and volume of each box is given in the chart below. Calculate the density of each box. Density = Mass/ volume which box
stira [4]

Box C will have the greatest density.

All boxes have the same volume.

Explanation:

We calculate the density using the following formula:

density = mass / volume

density of Box A = 10 g / 20 cm³ = 0.5 g/cm³

density of Box B = 30 g / 20 cm³ = 1.5 g/cm³

density of Box C = 170 g / 20 cm³ = 8.5 g/cm³

Box C will have the greatest density.

All boxes have the same volume.

Learn more about:

density

brainly.com/question/952755

#learnwithBrainly

7 0
2 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
2 years ago
Where are the electrons found in Bohr's atomic model?
Reptile [31]

Answer:

In the shell...

Explanation:

In the electronic shell, or the orbit..

Bohr's Atomic Model was given by Niels Bohr. He proposed that the electrons inside an atom moved around in orbits or shells. There were different shells in the atom named as K, L, M, N... These were also called energy levels...

4 0
3 years ago
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