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Mila [183]
3 years ago
5

Why can't we carry an umbrella during lighting?

Chemistry
1 answer:
sukhopar [10]3 years ago
6 0
Because the medal rod is a conductor and will electricity goes straight through it and will hit you with lightning
You might be interested in
Can you help me balance the equations from 1-10 please and thank you.
Korolek [52]

Answer:

  1. 2 NaBr + Ca(OH)2 → CaBr2 + 2NaOH
  2. 2 NH3 +H2SO4 → (NH4)2(SO4)
8 0
2 years ago
Please match the orbital type with the correct number of orbitals
saul85 [17]
Hi there!

p = e-3
s = f-1
f = i-7
d = g-5

Hope that helps!
Brady
7 0
3 years ago
Determine the heat given off to the surroundings when 9.0 g of aluminum (fm = 26.98) reacts according to the equation 2al + fe2o
Akimi4 [234]

Hey there!:

Number of moles:

Molar Mass Al = 26.98 g/mol

n = mass / molar mass

n = 9.0 / 26.98

n = 0.3336 moles of Al

Given the reaction :

2 Al + Fe2O3 = Al2O3 + 2 Fe

From the equation, 2 moles of Al give off 849 kJ of heat :

Actual heat given off :

0.3336 / 2 * 849 =

0.3336 / 1698 = 1.4*10² Kj


Hope that helps!


8 0
2 years ago
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

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

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?
Ira Lisetskai [31]

Answer:

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g. °C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

5 0
2 years ago
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