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Marat540 [252]
2 years ago
10

Atoms are the smallest particles of a compound that still retain the properties of that

Chemistry
1 answer:
N76 [4]2 years ago
5 0

Answer:

the answer is true

Explanation:

it is the smallest particle in an element that takes part in a chemical reaction

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20 moles of NH3 are needed to produce ? Moles of H2O
AfilCa [17]

Hi :)

20 mol NH3 x 6 H2O/4 NH3 = 30 mol H2O

Hope this helped :)

5 0
3 years ago
Galaxies are composed of many different objects. What kind ok objects make up most of the visible matter in a galaxy?
Arte-miy333 [17]
Stars make up most of the visible matter
5 0
2 years ago
Read 2 more answers
A sheet of gold weighing 8.8 g and at a temperature of 10.5°C is placed flat on a sheet of iron weighing 19.5 g and at a tempera
timofeeve [1]

Answer:

T = 36.393\,^{\textdegree}C

Explanation:

The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:

-Q_{iron} = Q_{gold}

-(0.008\,kg)\cdot (452\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-54.4\,^{\textdegree}C) = (0.0195\,kg)\cdot (129\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-10.5\,^{\textdegree}C)

-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)

-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C

-1.438\cdot T +78.227^{\textdegree}C = T - 10.5^{\textdegree}C

2.438\cdot T = 88.727\,^{\textdegree}C

The final temperature is:

T = 36.393\,^{\textdegree}C

8 0
3 years ago
Read 2 more answers
When hydrogen is burned in the presence of oxygen it will form water, as per the equation: 2H2 + O2 -> 2H2O. Which types of r
nadya68 [22]

Answer:

B

Explanation:

Hydrogen is synthesized to water by adding Oxygen.

Hydrogen is oxidised to water by combustion ( burning in presence of oxygen).

4 0
3 years ago
Consider the half reactions below for a chemical reaction.
ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0
2 years ago
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