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3 years ago

A permanent magnet can affect:

Chemistry

1 answer:

ruslelena [56]3 years ago

3 0

A. both permanent magnets and electromagnets.

Explanation:

A permanent magnet can affect and attract any other permanent magnet and even electromagnet.

They also affect any magnetic materials especially metals that can be magnetized.

In the vicinity of such substances, an attractive or repulsive force sets in and they both interact in the presence of the force field in place.

Permanent magnets cannot magnetize non-magnets.

An electromagnet is a magnet produced by the passage of electric current through a wire wound round a metallic core.

learn more:

Electromagnet brainly.com/question/2191993

#learnwithBrainly

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Zepler [3.9K]

The answer would be -3

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2 years ago

Read 2 more answers

Come up with 2 examples of derived units from everyday life.

kirill115 [55]

Answer:

answer

Explanation:

Area : Square Meters,

Volume : Cubic Meters,

Density : Kilograms per Meter Cube,

Velocity : Meters per Second Squared,

Force : Newtons (Mass multiplied by Acceleration or Mass multiplied by Displacement divided by Time squared)…..

5 0

2 years ago

The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca

geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

$M=d\times V=1.00 g/mL\times 150 ml =150.0 g$

Mass of solution = m = M = 150.0 g

$NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol$

Moles of NaOH = $\frac{5.00 g}{40 g/mol}=0.125 mol$

Energy released when 0.125 moles of NaOH added in water = Q

$Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J$

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

$Q'= 5,563.8 J$

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = $\Delta T$

$Q'=mc\times \Delta T$

$5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T$

$\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC$

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0

3 years ago

What is the mass of 0.73 moles of AgNO3?

love history [14]

Answer:

124 g (3 sig figs)

124.011 g (6 sig figs

Explanation:

Step 1: Calculate g/mol for AgNO₃

Ag - 107.868 g/mol

N - 14.01 g/mol

O - 16.00 g/mol

107.868 + 14.01 + 16.00(3) = 169.878 g/mol

Step 2: Multiply 0.73 moles by molar mass

0.73 mol (169.979 g/mol)

124 grams of AgNO₃

6 0

3 years ago

What kind of heat transfer happens in a microwave

valina [46]

Answer:

radition heat transfer in microwave

8 0

2 years ago