Does mass<span> alone provide no information about the amount or size of a measured quantity? No, we need combine </span>mass<span> and </span>volume<span> into "one equation" to </span>determine<span> "</span>density<span>" provides more ... </span>g/mL<span>. An </span>object has<span> a mass of </span>75 grams<span> and a volume of </span>25 cc<span>. ... A </span>certain object weighs 1.25 kg<span> and </span>has<span> a </span>density of<span> </span>5.00 g/<span>mL</span>
¹/3 C3H8(g) + ⁵/3 O2(g)
Explanation:
The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;
For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).
Therefore to produce 1.00 mole of CO₂(g);
We represent it in ratio;
C₃H₈(g) : CO₂(g)
1 : 3
What about ;
? (x) : 1
We cross multiply;
3x = 1 * 1
X = 1/3
We evaluate the same for O₂;
O₂(g) : CO₂(g)
5 : 3
What about
? (x) : 1
3x = 5 * 1
x = 5/3
Learn More:
For more on evaluating moles in chemical reactions check out;
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They both perform photosynthesis and also cell respiration
Answer:
The answer is hardness and brittleness. The ion discharging procedure and taking after impact on reservoir water quality file, for example, hardness, pH, conductivity, has been dissected. Comes about uncovered that the most discharge quality of various materials was touching base at the 30s after startup.
Explanation:
Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)
