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Nuetrik [128]
3 years ago
8

Some magnesium powder was mixed with some copper( II)oxide and heated strongly. there was a victorious reaction producing a lot

of sparks and a bright flash of light.
1) name the products of the reaction
2)write a balanced equation for the reaction.

plzzzzz help!!!​
Chemistry
2 answers:
Ilya [14]3 years ago
8 0

Answer:

See below.

Explanation:

1)  Magnesium oxide MgO  and copper metal (Cu).

2) Mg + CuO --->  MgO + Cu.

-BARSIC- [3]3 years ago
4 0

Reaction:

\mathrm{Mg \: + \: CuO \rightarrow Cu\: + \: MgO}

Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.

Products of the reaction: Magnesium oxide and metallic copper.

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A catalyst speeds up the rate of reaction so the answer is B.
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3 years ago
The ion with 36 electrons, 34 protons, and 44 neutrons
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7 0
3 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
If carbon-15 has a half-life of 2.5 seconds, % of the sample would still be
kramer

Answer:

After 5 second 25% C-15 will remain.

Explanation:

Given data:

Half life of C-15 = 2.5 sec

Original amount = 100%

Sample remain after 5 sec = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 5 sec / 2.5 sec

Number of half lives = 2

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

Thus after 5 second 25% C-15 will remain.

8 0
4 years ago
C4H10+ 02 → _CO2 + __ H20
Svetllana [295]

Answer:a) 2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction

b) 8 (4 moles CO2 per mole butane)

Explanation:

could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O

4 0
3 years ago
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