Answer:
Answers
1.)reactants: nitrogen and hydrogen; product: ammonia.
2.)reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water.
3.)N 2 + 3H 2 → 2NH 3
4.)Mg(OH) 2 + 2HNO 3 → Mg(NO 3) 2 + 2H 2O.
5.)2NaClO 3 → 2NaCl + 3O 2
6.)4Al + 3O 2 → 2Al 2O 3
7.)N 2(g) + 3H 2(g) → 2NH 3(g)
Explanation:
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Answer:
frequency = 0.47×10⁴ Hz
Explanation:
Given data:
Wavelength of wave = 6.4× 10⁴ m
Frequency of wave = ?
Solution:
Formula:
Speed of wave = wavelength × frequency
Speed of wave = 3 × 10⁸ m/s
Now we will put the values in formula.
3 × 10⁸ m/s = 6.4× 10⁴ m × frequency
frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m
frequency = 0.47×10⁴ /s
s⁻¹ = Hz
frequency = 0.47×10⁴ Hz
Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.
Answer:
Mixtures
Explanation:
Matter can be classified as a compound and a mixture.
Answer : The excess reactant in the combustion of methane in opem atmosphere is 
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
= 
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give 
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
C. single replacement. A single replacement is represented by this formula : AB + C = AC + B
