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3 years ago

A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is:

2 answers:

7 0

Its magnifying power is: 4X 5X 9X 20X. A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is 9x. This answer has been confirmed as correct and helpful.

podryga [215]3 years ago

7 0

20x

explanation? I dont need to give one

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A. Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculat

liraira [26]

Answer:

A)Mass of gallium plated out is 0.3440 grams

B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Explanation:

To calculate the total charge, we use the equation:

$C=I\times t$

where,

C = Charge

I = Current in time t (seconds)

To calculate the moles of electrons, we use the equation:

$\text{Moles of electrons}=\frac{C}{F}$

where,

F = Faraday's constant = 96500

A) The equation for the deposition of Ga(s) from Ga(III) solution follows:

$Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)$

I = 0.790 A, t = 30.0 min = 1800 seconds

$C=I\times t$

$C=0.790 A\times 1800 s=1422 C$

Moles of electron transferred:

$=\frac{1422 C}{96500 F}=0.01474 mol$

Now, to calculate the moles of gallium, we use the equation:

$\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}$

n = number of electrons transferred = 3

$\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol$

Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g

B) The equation for the deposition of Sn(s) from Sn(II) solution follows:

$Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$

Moles of tin = $\frac{8.70 g}{119 g/mol}=0.07311 mol$

n = number of electrons transferred = 2

$\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}$

Moles of electron = $n\times \text{Moles of tin}$

$=2\times 0.07311 mol=0.14622 mol$

Charge transferred during time t :

$\text{Moles of electrons}=\frac{C}{F}$

$C=96500 F\times 0.14622 mol=14,110.23 C$

Current applied for t time = I = 5.79 A

$t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs$

For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

5 0

4 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

Acids react with what?

enot [183]

Answer:

Explanation:

The answer is A

7 0

3 years ago

Read 2 more answers

You have 47.0 mL of a 2.00 M concentrated or "stock" solution that must be diluted to 0.500 M. How much water should you add?

Inessa [10]

The required volume of water to make the dilute solution of 0.5 M is 188 mL.

<h3>How do we calculate the required volume?</h3>

Required volume of water to dilute the stock solution will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of stock solution.
M₂ & V₂ are the molarity and volume of dilute solution.

On putting values from the question to the above equation, we get

V₂ = (2)(47) / (0.5) = 188mL

Hence required volume of water is 188 mL.

To know more about volume & concentration, visit the below link:
brainly.com/question/7208546

#SPJ1

6 0

2 years ago

Look at the reaction below. H₂SO₄(aq) + Ca(OH)₂(aq) ---> CaSO₄(aq) + 2H₂O(/) Which substance is the base in the reaction?

Lelu [443]

Ca(OH)2 (aq) because of elements oxygen and hydrogen

4 0

3 years ago

Read 2 more answers