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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 the balenced equation of Cu+H2SO4---> CuSO4+H2O+SO2
algol [13]
Balanced chemical equation :

1 Cu + 2 H2SO4 = 1 CuSO4 + 2 H2O + 1 SO<span>2</span>

hope this helps!.

8 0
3 years ago
Mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has dissolved. At what temperature does this occur?
lozanna [386]

240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.

Explanation:

As per the solubility curve graph of different salts dissolved in 100 g of water, it can be noted that the KClO₃ gets saturated at 100 °C with the solubility of 59 g in it. But here the water is taken as 300 g and the salt is taken as 120 g. The graph shows the solubility of grams of salt in 100 g of water.

So, then we have to find what is the gram of salt present in 100 g of water for the present sample.

300 g of water contains 120 grams of salt.

Then, 1 g of water will contain \frac{120}{300} g of salt.

Then, 100 g of water will contain \frac{120}{300}*100 of salt. This means, 100 g of water will contain 40 g of salt in it.

Then, if we check the graph, the corresponding x-axis point on the curve for the y-axis value of 40 g will give us the temperature required to dissolve the salt in water. So 80°C will be required to dissolve 40 g of salt in 100 g of water.

Then, in order to get the temperature required to dissolve 120 g in 300 g of water, multiply the temperature by 3, which will give the result as 80×3=240 °C.

So, 240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.

8 0
3 years ago
How are bacteria and humans different?
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7 0
3 years ago
Is baking powder Litmus acid or base
liubo4ka [24]

Answer:

It is a base and should turn a paper green

Explanation:

6 0
2 years ago
An aluminum kettle weights 1.05 kg and has a heat capacity of 0.9211 J over grams Celsius how much heat is required to increase
Serggg [28]

Answer:

64799.4 J

Explanation:

The following data were obtained from the question:

Mass (M) = 1.05 kg = 1.05 x 1000 = 1050g

Specific heat capacity (C) = 0.9211 J/g°C

Initial temperature (T1) = 23°C

Final temperature (T2) = 90°C

Change in temperature (ΔT) = T2 – T1 =

90°C – 23°C = 67°C

Heat required (Q) =....?

The heat required to increase the temperature of the kettle can b obtain as follow:

Q = MCΔT

Q = 1050 x 0.9211 x 67

Q = 64799.4 J

Therefore, 64799.4 J of heat is required to increase th temperature of the kettle from 23°C to 90°C.

4 0
3 years ago
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