Honestly I don’t even know
Answer:
2L of nitrogen gas will be needed
Explanation:
Based on the following reaction:
N₂ + 3H₂ → 2NH₃
<em>1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.</em>
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If 6L of hydrogen (In a gas, the volume is directly proportional to the moles, Avogadro's law) react, the volume of nitrogen gas required will be:
6L H₂ * (1mol N₂ / 3 moles H₂) =
<h3>2L of nitrogen gas will be needed</h3>
Explanation:
Light intensity influences the manufacture of plant food, stem length, leaf color and flowering. Generally speaking, plants grown in low light tend to be spindly with light green leaves. A similar plant grown in very bright light tends to be shorter, better branches, and have larger, dark green leaves.
Answer: Option (C) is the correct answer.
Explanation:
An ion is defined as a specie which is formed when a neutral atoms tends to gain or lose an electron.
When a neutral atom gain an electron then it forms a negative ion whereas when a neutral atom tends to lose an electron then it forms a positive ion.
For example, a neutral fluorine atom on gaining an electron will form
ion. And, a sodium atom on losing an electron forms
ion.
When a group of atoms form ions then it tends to form polyatomic ions.
Thus, we can conclude that group of atoms that gains or loses electrons is called a polyatomic ion.
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
