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blsea [12.9K]
3 years ago
11

This chemical equation represents a ______________ reaction.

Physics
2 answers:
Ede4ka [16]3 years ago
7 0

Answer:

you didnt put anything

Explanation:

BARSIC [14]3 years ago
3 0
Yess yess it’s correct !!!!
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If the current flowing through an electric heater increases from 6 to 12 amp while the voltage remains the same, the heat produc
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Q=l^2 times R times t
Where Q - heat, I -current, R - resistance and t is time

If you increase I twice (and it's squared), than Q gonna went up 4 times (2 squared).

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4 years ago
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G of potassium reacts with 16 g of oxygen to produce 94 g of potassium oxide
Vanyuwa [196]

Answer:

78g

Explanation:

Given parameters:

Mass of oxygen gas = 16g

Mass of potassium oxide = 94g

Unknown:

Mass of reacting potassium = ?

Solution:

To solve this problem, we need to obtain a balanced reaction equation. Then determine the number of moles of the reactant and use it to find that of the other one.

Balanced equation:

                      4K +   O₂   →    2K₂O

  Number of moles of reacting oxygen;

       Number of moles = \frac{mass}{molar mass}

molar mass of  O₂ = 2 x 16 = 32g/mole

     Number of moles = \frac{16}{32}  = 0.5mole

From the reaction equation;

          4 mole of K reacted with 1 mole of O₂;

           x mole of K will react with 0.5 mole of O₂

Therefore, 4 x 0.5 = 2 moles of potassium.

Mass of potassium = number of moles x molar mass

              Molar mass of potassium = 39g

Mass of potassium = 2 x 39 = 78g

5 0
3 years ago
Early in Earth’s history, differentiation resulted in ___________.
ICE Princess25 [194]

Answer:

ITS D

Explanation:

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3 years ago
A planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbit
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The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:
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where
r_1 is the orbital radius of the first satellite
r_2 is the orbital radius of the second satellite
T_1 is the orbital period of the first satellite
T_2 is the orbital period of the second satellite

If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
T_2 =  \sqrt{T_1^2 ( \frac{r_2}{r_1} )^3} = \sqrt{(18.0 h)^2 ( \frac{3\cdot 10^7 m}{2 \cdot 10^7 m} )^3}  = 33.1 h
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What statement is true about a spring?
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A spring is felxiable
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