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serg [7]
3 years ago
12

If 5.0 liters of carbon dioxide gas are produced by the reaction below at STP, how many liters of oxygen gas were used in the re

action?
2CO (g) + O2 (g) 2CO2 (g)
10.0 L
7.5 L
5.0 L
2.5 L
Chemistry
1 answer:
svp [43]3 years ago
4 0
At STP

1L O₂ → 2L CO₂
xL O₂ → 5.0L CO₂

x=2.5 L
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What causes the emission of radiant energy that produces characteristic spectral lines for a given element?​
Anarel [89]

Answer:

When electrons move from a higher energy level to a lower one, photons are emitted, and an emission line can be seen in the spectrum. Absorption lines are seen when electrons absorb photons and move to higher energy levels. ... An atom in its lowest energy level is in the ground state.

6 0
3 years ago
Read 2 more answers
What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound.  That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms.  The empirical formula is therefore N₂O.

8 0
3 years ago
Una solución formada por 58 gramos de cloruro de sodio (NaCl) en 100 mL de solución posee una concentración
masya89 [10]

Answer:

58

Explanation:

8 0
3 years ago
How many significant figures<br> are in this number?<br> 43.55
leva [86]

Answer:

4 significant figures

Explanation:

Significant figures are the units/digits within a number that make the number more accurate and precise.

All digits (except for 0) are always significant. Therefore, all the digits in 43.55 are significant. Since there are 4 digits in the given number, there are 4 significant figures.

7 0
2 years ago
Hydrazine, n2h4, is used as a rocket fuel. in the reaction below, if 80.1 g of n2h4 and 92.0 g of n2o4 are allowed to react, how
Mama L [17]
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.

6 0
4 years ago
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