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vazorg [7]
4 years ago
10

when a substance donates a proton during a chemical reaction, that substance can be classified as a(an) ___ ?

Chemistry
1 answer:
Anarel [89]4 years ago
4 0

Answer:

An acid

Explanation:

According to the Arrhenius theory of acids and bases "an Arrhenius acid is a compound that increases the concentration of H+ ions that are present when added to water"

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The particles of a gas are<br> O electrons<br> O atoms or molecules<br> O neutrons<br> O waves
Harrizon [31]
Answer:
atoms or molecules

Explanation:
Gas particles are constantly bumping into each other and the borders of their container.

8 0
2 years ago
Tonya had a mixture of sand, salt, iron particles, and pebbles. What processes should be used to separate the substances in the
marusya05 [52]

Explanation:

Put the pan into a water-containing dish and mix well. Now use a strainer to transfer the solution into another jar. The salt should disappear in it. 

And using a tube with a filtrate, transfer the salts that has sand into another bottle with a filtrate. Therefore the sand is split. Eventually, when all the water vaporizes and the salt stays in the bottle, leave the extra solvent and heat it.

3 0
3 years ago
Why are numbers placed before the elements and molecules in a<br>chemical equation?​
Neko [114]

Answer:

the are coefficients for the atom. the tell how many there are.

Explanation:

they tell you how many atoms there are of its type.

ex. 4 C²H³. there are 4 of the c² and h³ (couldn't find a way to use subscript so used the power signs to show.)

7 0
3 years ago
What are thermal plasmas?
vitfil [10]
<span>plasmas that reach a temperature equal to their surroundings




----------------------=0uyjgyhjngytfgbhtfgbdc b gdf gb </span>
6 0
3 years ago
Read 2 more answers
The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does t
Mama L [17]

Answer:

6.1%

Explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law  at constant pressure,

\frac{V_1}{T_1} =\frac{V_2}{T_2}

T_2=39\°\ C=39+273=312\ K

T_1=59°\ C=273+59=332\ K

V_2=V_1\times \frac{T_2}{T_1} \\=V_1\times\frac{312}{332}\\=0.939V_1

Percentage change in volume

\%\ change\ in\ volume=\frac{V_1-0.939V_1}{V_1} \times 100\\=6.1\%

Change in volume of the balloon is 6.1%

4 0
3 years ago
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