Is a compound microscope a complex microscope??? Give your reason why it is :P

If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density

ankoles [38]

Answer:

The length of the wire = 352.66 feet.

Explanation:

A copper refinery produces a copper ingot weighing 150 lb. If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3. (Assume that the wire is a cylinder whose volume is V = πr2h, where r is the radius and h is its height or length.)

Step 1: Convert lb to kg

150 lb = 68.0389 kg

Step 2: Calculate volume of copper

Volume = mass / density

Volume = 68038.9 grams / 8.94 g/cm³

Volume = 7610.6 cm³ Cu

Step 3: Calculate length of wire

The diameter of the wire is 9.50 mm, so the radius is half of that (4.75 mm), or 0.475 cm.

The total "volume" of the wire is πr²h = (π)*(0.475 cm)²(h) = 0.708h = 7610 cm^3

7610 = 0.708h

h = 10749 cm = length of wire

The length of the wire = 352.66 feet.

7 0

3 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

1. How many molecules of S2 gas are in 756.2 L?

AfilCa [17]

Answer: There are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

Explanation:

It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of moles present in 756.2 L are calculated as follows.

$Mole = \frac{Volume}{22.4 L}\\= \frac{756.2 L}{22.4 L}\\= 33.76 mol$

According to mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

$1 mol = 6.022 \times 10^{23}\\33.76 mol = 33.76 \times 6.022 \times 10^{23}\\= 2.032 \times 10^{25}$

Thus, we can conclude that there are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

5 0

3 years ago

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is

Reika [66]

Answer: The pH of the buffer is 4.61

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.70

$[\text{conjuagate base}]}$ = moles of conjugate base = 3.25 moles

$[\text{acid}]$ = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

$pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61$

Hence, the pH of the buffer is 4.61

8 0

3 years ago

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻ (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

Learn more here:

brainly.com/question/10547418?referrer=searchResults

brainly.com/question/14096111?referrer=searchResults

I hope it helps you!

3 0

2 years ago