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4 years ago

Lamp, toaster, and a fan

Chemistry

1 answer:

Usimov [2.4K]4 years ago

5 0

Answer: Light, Thermal, Kinetic

Explanation:

The lamp converts the electric energy to light energy.

The toaster converts the electric energy to thermal energy.

The fan converts the electric energy to kinetic energy.

Good luck with your test!

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When 10.0 grams of ch4 reacts completely with 40.0 grams of o2 such that there are no reactants left over, 27.5 grams of carbon

Zina [86]

The balanced equation for the reaction is :-

CH₄(g) + 2O₂(g) ---------> CO₂(g) + 2H₂O(l)

Molar mass of CH₄ = 16 g/mole

Molar mass of O₂ = 32 g/mole

Molar mass of H₂O = 18 g/mole

Molar mass of CO₂ = 44 g/mole

Now we calculate the number of moles of reactants,

moles of CH₄ = mass of CH₄/molar mass of CH₄ = 10/16 = 0.625

Moles of O₂ = mass of O₂/molar mass of O₂ = 40/32 = 1.25

Now, as per the balanced reaction, for complete reaction to occur, one mole of CH₄ require 2 moles of O₂

Thus, 0.625 moles of CH₄ requires 1.25 moles of O₂

both CH₄ and O₂ are present in the exact required quantity.

Hence, moles of CO₂ formed = moles of CH₄ reacted = 0.625

Mass of CO₂ formed = moles of CO₂ x Molar mass of CO₂ = 0.625 x 44 = 27.5g

Moles of H₂O formed = moles of O₂ reacted = 1.25g

Thus, mass of H₂O formed = moles of H₂O x molar mass of H₂O = 1.25 x 18 = 22.5g

8 0

3 years ago

The unbalanced equation below shows the reaction between sodium and water. In the reaction,

NISA [10]

Answer:

41.4 g

Explanation:

2Na(s) + 2H2O (l) ----> 2NaOH ( aq) + H2 (g)

According to the above balnced chemical equation the ratio between moles of H2 gas formed to moles of Na used up is 1:2

Molar mass ( Mass of 1 mol) Of H2 gas is 2 gmol-1

So in 1.8 g of H2 0.9 mol should be present

Therefore no. of moles of Na used should be 2× 0.9 = 1.8 moles

Molar mass of Na= 23 gmol-1

So mass of 1.8 mol of Na = 1.8× 23 g = 41.4g

5 0

3 years ago

When nitrogen gas reacts with hydrogen gas, ammonia gas is formed. how many grams of hydrogen gas are required to react complete

aniked [119]

The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required

3 0

4 years ago

The catalytic decomposition of hydrogen peroxide yields oxygen gas and water, according to the reaction given below. In an exper

elixir [45]

Answer:

The molar concentration of hydrogen peroxide is 0.01164 M

Explanation:

Step 1: Data given

Volume of gas yielded = 75.3 mL = 0.0753 L

Temperature = 25.0 °C

Atmosphere = 0.976 atm

(Water = 0.032 atm at 25°C)

The original volume of H2O2 is 500 mL

Molar mass of O2 =32 g/mol

Step 2: The balanced equation

2H2O2(aq) → 2 H2O(l) + O2(g)

Step 3: Calculate pressure of O2

P = P(02) + P(water)

P(O2) = P - P(water)

P(O2) = 0.976 atm - 0.032 atm

P(O2) = 0.944 atm

Step 4: Calculate moles O2

p*V=n*R*T

⇒ with p = the pressure of O2 gas = 0.944 atm

⇒ with V= the volume of O2 = 0.0753 L

⇒ with n = the number of moles of O2

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.944 * 0.0753)/(0.08206*298K)

n = 0.00291 moles O2

Step 5: Calculate moles of H2O2

For 1 mole of O2 produced, we need 2 moles of H2O2

For 0.00291 moles O2 we need 2*0.00291 = 0.00582 moles H2O2

Step 6: Calculate molar concentration of H2O2

Molar concentration = moles / volume

Molar concentration = 0.00582 moles / 0.500L

Molar concentration = 0.01164 M

The molar concentration of hydrogen peroxide is 0.01164 M

3 0

3 years ago

What would a Bohr diagram look like for Oxygen?

stealth61 [152]

Under normal circumstances, it's an "S" shape. The graph below shows both a normal Bhor diagram and one during exercise or what otherwise would cause a high release of Carbon dioxide. Oxygen dissociates much more readily and easily if the concentration of CO2 is higher.

6 0

3 years ago