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3 years ago

Look at the data table and graph for the 25-coil electromagnet. With the 1.5 V battery, the

1 answer:

4 0

Answer a:
A 1.5 V battery, the electromagnet picked up an average of 6 paper clips, while with the 6.0 V battery, an average of 23 paper clips were picked up. Battery of 6.0V is 6.0/1.5 = 4 times stronger than battery of 1.5 V

Answer b:
Ratio of the number of paper clips picked up using the 6.0 V battery to the number picked up using the 1.5 V battery is = 23/6 = 3.8 ≈ 4.

Answer c:
As the voltage power increase, more paper clips were picked up by electromagnet. This indicated that there is a direct relationship. Mathematically it can be expressed as:

Voltage Power α Number of paper clips that were picked up

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Explain how you can use an atoms mass number and atoic number to determine the number of protrons electrons and neutrons in the

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Simple!

Atomic number= number of protons AND number of electrons (in an atom)

Mass number= number of neutrons

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3 years ago

What is Gibbs free energy

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Answer:

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Explanation:

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1 If gibbs free energy change of a reaction is negative then the reaction is spontaneous.

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3 If free energy change is positive then the reaction is non spontaneous.

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3 years ago

Read 2 more answers

Write packing efficiency of fcc ,BCC, SCC and formula also . ?

Minchanka [31]

❖ Packing Efficiency ❖

➪ The percentage of total space occupied by particles is called packing efficiency.

$\\ \qquad{\rule{200pt}{3pt}}$

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

❖ Packing efficiency of simple cubic structure (SCC).❖

$\rm \: {Let \: } \bf{r } \: \rm{ be \: the \: radius \: of \: a \: sphere }\: \\ \rm and \: \bf{a} \: \rm {be \: the \: edge \: of \: unit \: cell.} \\ \\ \; \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar$

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

$: : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2r)³

Volume of unit cell = 8r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪ Substituting the known values in the formula, we get the following results:

$: : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ : : \implies \dfrac{\pi}{6} \times 100 \\\\\ : : \implies \bf {52.4 \%}$

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

$\\ \qquad{\rule{200pt}{3pt}}$

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\ \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\ \implies\bf52.4 \%$

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

$\\ \qquad{\rule{200pt}{3pt}}$

❖ Packing efficiency of body cubic structure (BCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\ \implies \dfrac{3\pi}{8} \times 100 \\\ \\ \bf \implies \: 68 \%$

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

$\\ \qquad{\rule{200pt}{3pt}}$

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2 years ago

What mass of silver chloride can be produced from 1.30 L of a 0.245 M solution of silver nitrate? The reaction described in Part

Gnoma [55]

Answer:

Explanation:

Given parameters:

Volume of AgNO₃ = 1.3L

Molarity or concentration AgNO₃ = 0.245M

Unknown:

Mass of AgCl produced = ?

Solution:

To solve this problem, we have to work from the known compound to the unknown one using the mole concept.

The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;

Now let us establish the balanced reaction equation;

2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂

Now;

Find the number of moles of AgNO₃;

Number of moles of AgNO₃ = concentration x volume

= 1.3 x 0.245

= 0.32moles

From the balanced reaction equation;

2 mole of AgNO₃ produced 2 moles of AgCl

0.32 moles of AgNO₃ will produce 0.32moles of AgCl

Now that we know the number of moles of the AgCl, we can find the mass;

Mass of AgCl = number of moles x molar mass

Molar mass of AgCl = 107.9 + 35.5 = 143.4g/mol

Mass of AgCl = 0.32 x 143.4 = 45.89g

5 0

3 years ago