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Minchanka [31]
4 years ago
5

What is the difference between o2 and f2

Chemistry
1 answer:
Flura [38]4 years ago
5 0
<span>O2 and F2 are both diatomic in nature and both exist as gas at normal temperature and pressure. Since Fluorine gas is formed from noble element fluorine it is less reactive while Oxygen is more reactive than fluorine gas and the same conditions.</span>
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When a student combed his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened? A) The
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B. 

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3 years ago
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5.77 g of nitrogen react with excess hydrogen producing 6.83 g of ammonia what is the percent yield. Determine/Label the actual
nata0808 [166]

Answer:

Percentage yield is 41.21%

Explanation:

Equation of reaction,

N₂ + 3H₂ → 2NH₃

Actual NH3 = 6.83g

Mass of N2 = 5.77g

Theoretical yield = ?

5.77g of N2 = 6.83g of NH3

14g of N2 = xg

X = (14 × 6.83) / 5.77

X = 95.62 / 5.77

X = 16.57g of NH3

Theoretical yield of NH3 is 16.57g

Percentage yield = (actual yield / theoretical yield) × 100

% yield = (6.83 / 16.57) × 100

% yield = 0.4121 × 100

% yield = 41.21%

The percentage yield of NH3 is 41.21%

6 0
3 years ago
The radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-g sample of radon- (3 222 would be left after approxi
Novay_Z [31]

Answer:

1.115 g

Explanation:

Applying,

R = R'2^{n/t}................... Equation 1

Where R = original sample of radon-222, R' = sample of radon-222 left after decay, n = Total time, t = half-life.

make R' the subject of the equation

R' = R/(2^{n/t})............... Equation 2

From the question,

Given: R = 73.9 g, n = 23 days, t = 3.8 days.

Substitute these values into equation 2

R' = 73.9/(2^{23/3.8})

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R' = 1.115 g

6 0
3 years ago
Which statement describes the best way to determine how different levels of light affect the growth of seedling plants?
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Choose three different light levels, and place four identical plants under each light level to observe the light’s effect on multiple plants.

Explanation:

4 0
3 years ago
2Al+6HBr -&gt; 2AlBr3+3H2. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?
Cerrena [4.2K]
<span>2 Al+6 HBr =  2 AlBr</span>₃ <span>+ 3 H</span>₂

2 moles Al ---------  6 moles HBr ----------- 3 moles H₂
3.22 moles Al ------ 4.96 moles HBr ----- ( moles H₂ )

moles H₂ = 4.96 x 3 / 6

moles H₂ = 14.88 / 6

= 2.48 moles of H₂

hope this helps!
4 0
3 years ago
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