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Xelga [282]
3 years ago
7

Do hydrogen lose 1 2 or 3 electrons to other atoms

Chemistry
1 answer:
iren2701 [21]3 years ago
7 0

Answer:<em> Hydrogen can lose as much as possible there is no limits to it.</em>

<em>Hope this helps!</em>

<em>I am joyous to assist you anytime!</em>

<em>-Jarvis</em>

<em>Extras: Hydrogen is the chemical element with the symbol H and atomic number 1. hydrogen is the lightest element in the periodic table. Hydrogen is the most abundant chemical substance in the Universe (;</em>

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When heated, calcium hydroxide and ammonium chloride react to produce ammonia gas, water vapor, and solid calcium chloride.
NNADVOKAT [17]

Answer:

13.73g

Explanation:

mass of reactants = mass of products.

Mass reactants = 5.00 g + 10.00 g = 15.00 g

Mass products = 1.27g + mass of ammonia and water vapor

Mass of ammonia and water vapor

15.00g – 1.27 g = 13.73 g

3 0
3 years ago
How do u circulate netforce
emmasim [6.3K]

1Draw a quick sketch of the object.

2Draw an arrow showing every force acting on the object.

3To calculate the net force, add any vectors acting on the same axis (x and y), making sure to pay attention to the directions.

5 0
3 years ago
A 49.7 mL sample of gas in the cylinder is warm from 20°C to 92°C. What is the volume of the final temperature
Xelga [282]

Answer:

61.9mL

Explanation:

Given parameters:

Initial volume  = 49.7mL

Initial temperature  = 20°C  = 293K

Final temperature  = 92°C  = 365K

Unknown:

Final volume  = ?

Solution:

To solve this problem, we simply apply the Charles's law which states that "the volume of a fixed mass of a gas varies directly as it absolute temperature if the pressure is  constant";

        \frac{V1}{T1}   = \frac{V2}{T2}  

V and T are volume and temperature values

1 and 2 are initial and final states

 Now insert the parameters and solve;

      \frac{49.7}{293}   = \frac{V2}{365}  

        V2  = 61.9mL

6 0
3 years ago
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 11.0 g of aluminum
dlinn [17]

Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of aluminium}=\frac{11.0g}{27g/mol}=0.407moles

2Al+3Cl_2\rightarrow 2AlCl_3  

According to stoichiometry :

Al is the limiting reagent as it limits the formation of product and Cl_2 is the excess reagent.

As 2 moles of Al give = 2 moles of AlCl_3

Thus moles of Al give =\frac{2}{2}\times 0.407=0.407moles  of AlCl_3

Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

5 0
3 years ago
The equation 2NO2 ------N2O represent a system_______
Karolina [17]

Answer:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

Explanation:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

6 0
2 years ago
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