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wlad13 [49]
2 years ago
7

For the following chemical equation.

Chemistry
1 answer:
kotegsom [21]2 years ago
5 0

Sr is the limiting reactant.

Given the reaction equation;

2Sr + O2 (g) → 2SrO

2 moles of Sr reacts with 1 mole of O2

2 moles Sr will react with x mole of O2

x = 2 ×1/2

x = 1 mole of O2

Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.

Learn more: brainly.com/question/14225536

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Why cant (NaCl) light the bulb until the NaCl is dissolved in water. *
Aneli [31]

Answer:

NaCl will only conduct electricity in solutions

Explanation:

For electrical conduction, free mobile electrons as seen in most metals must be present or ions which are charged particles must be available for solutions and molten substances.

  • Sodium chloride is an ionic compound without free mobile electrons or ions despite being ionic.
  • It will maintain a subtle and unique charge stability when in solid form.
  • In solid, the ions are not free to move and remain locked up in the solid mass.
  • When introduced into a solution, the ions becomes free to move and this will aid electrical conduction.
6 0
3 years ago
HELPPPP STILLLL!!!! SCIENCE!!!!
Vadim26 [7]

Answer: the third one

Explanation:

4 0
3 years ago
If 42.9 mL of rubbing alcohol is
lisov135 [29]

Answer:

19.8 %

Explanation:

V% = ( V of solute \ 100 ml  of solution ) ×100%  =

   the volume of the solute in 100ml solition = 20 ml

  42.9 ml of solute ----------215 ml solution

     ×                     -----------  100 ml solution

42.6 ×   100 ml  ÷ 215 ml

= 19.8 ml of solute

the V% =   (19.8 ÷ 100)  ×100% = 19.8%

3 0
3 years ago
Give three examples of monoatomic molecules.
umka2103 [35]

Answer:

Helium

Radon

and Neon

Explanation:

5 0
3 years ago
What mass of water could be warmed from 21.4 degrees celsius to 43.4 degrees celsius by the pellet dropped inside it? Heat capac
Artist 52 [7]

42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it

Heat loss by the pellet is equal to the Heat gained by the water.

q_{w} = -q_{p} ….(1)

where, q_{w} is the heat gained by water

q_{p} is the heat loss by pellet

q_{w} = mCΔT

where m = mass of water

C = specific heat capacity of water = 4.184 J/g-°C

ΔT = Increase in temperature

ΔT for water = 43.4 - 21.4 = 22°C

q_{w} = m × 4.184 × 22 …. (2)

Now

q_{p} = H_{c} ×ΔT

where H_{c} = Heat capacity of pellet = 56J/°C

Δ T for pellet = 43.4 - 113 =- 69.6°C

q_{p} = 56 × -69.6 = -3897.6 J

From equation (1) and (2)

-m× 4.184 × 22 =-3897.6

m= 42.34 g

Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.

Learn more about specific heat here brainly.com/question/16559442

#SPJ1

6 0
2 years ago
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