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3 years ago

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)

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Use Gay-Lussac Law to solve. When the pressure of a balloon starts at 7.5 atm and the temperature surrounding the balloon goes f

ElenaW [278]

Answer:

The answer to your question is 6.77 atm

Explanation:

Data

Pressure 1 = P1 = 7.5 atm

Temperature 1 = T1 = 65°C

Pressure 2 = P2 = ?

Temperature 2 = T2 = 32°C

Process

-Use Gay-Lussac law to solve this problem

P1/T1 = P2/T2

-Solve for P2

P2 = P1T2 / T1

-Convert temperature to °K

T1 = 65 + 273 = 338°K

T2 = 32 + 273 = 305°K

-Substitution

P2 = (7.5 x 305) / 338

-Simplification

P2 = 2287.5 / 338

-Result

P2 = 6.77 atm

6 0

3 years ago

How many moles are in 35.6 g of H2O

klasskru [66]

Answer:

Given

mass of H2O (m) =35.6g

molarmass (mr) = H2O ), 1x2+16=18g/mol

moles of H2O (n) =?

sln

n=m/mr

n=35.6g/18g/mol

n=1.978moles

the moles of H2O are 1.978moles

3 0

2 years ago

What volume of 0.205 m k3po4 solution is necessary to completely react with 154 ml of 0.0110 m nicl2? 1.65 l?

sasho [114]

The balanced equation for the above reaction is
2K₃PO₄ + 3NiCl₂ ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄ to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³ = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL

3 0

3 years ago

The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 218 kJ/mol and the change in enthalpy for the

madreJ [45]

Answer:

$470\frac{KJ}{mol}$

Explanation:

The activation energy represents the energy barrier that reagents must pass to transform into products (or products to transform into reagents in a reverse reaction)

For any reaction, the change in enthalpy is related to the activation energy by the equation

$\Delta H =E_{a\ direct}-E_{a\ reverse}$

So, the activation energy for the reverse reaction is

$E_{a\ reverse}=E_{a\ direct}-\Delta H =218 \frac{KJ}{mol} - (-252)\frac{KJ}{mol}=470\frac{KJ}{mol}$

4 0

3 years ago

1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te

love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

$Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole$

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0

3 years ago