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Gnesinka [82]
3 years ago
8

Describe the three types of magnets

Chemistry
1 answer:
Semmy [17]3 years ago
8 0

Answer:

Permanent Magnets are always magnetic.

Temporary Magnets loose their magnetizing power after the source is removed. (outside a magnetic field)

Electromagnets are made by sending electric coil around an iron rod. (I.e. Nail, coil, and batteries can pick up paperclips)

You might be interested in
The answer to (3.540)x(0.0065)x(401) should have ____.
miv72 [106K]
B. Two significant figures


Your final answer can only have as many significant figures as the number that has the least amount of significant figures in it, which is (0.0065)
3 0
3 years ago
If a piece of aluminum with a mass of 3.99 g and a temperature of 100.0 °C is dropped
Vinil7 [7]

Answer:

The final temperature of the system is 27.3°C.

Explanation:

Heat lost by aluminum = 3.99 × 0.91 × (100-T)

                                     = 3.631 (100-T)

Heat gained by water = 10 × 4.184 × (T-21)

                                    = 41.84 (T-21)

As,

                                Heat gained = Heat loss

                          or, 3.631(100-T) = 41.84(T-21)

                          or,363.1 -  3.631 T = 41.84 T - 878.64)

                          or, (41.84+ 3.631) T = 878.64 +363.1

                          or  T= \frac{1241.74}{45.47}

                         or, T = 27.3°C

Hence the final temperature is 27.3°C.

8 0
3 years ago
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
3 years ago
PLS HELP THE QUESTION IS ON THE PICTURE
IceJOKER [234]

<u>Concepts used:</u>

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: <em>Number of formula units = number of moles * 6.022*10²³</em>

number of moles of an element or a compound = given mass/molar mass

<u>__________________________________________________________</u>

<u>003 - </u><u>Number of CaH₂ formula units in 6.065 grams</u>

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

<u>004 </u><u>- Mass of 6.34 * 10²⁴ formula units of NaBF₄</u>

Number of Moles:

We mentioned this formula before:

<em>Number of formula units = number of moles * 6.022*10²³</em>

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles=  10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

<u>005</u><u> - Number of moles in 9.78 * 10²¹ formula units of CeI₃</u>

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles   OR   0.016 moles

3 0
3 years ago
PLEASE HELP ITS DUE TODAY
dimulka [17.4K]

Answer:

2nd one and last one

Explanation:

im just that smart

8 0
3 years ago
Read 2 more answers
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