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vova2212 [387]
3 years ago
15

Determine the ratio of the diatomic element with equal percentage abundance​

Chemistry
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

<em>I</em><em> </em><em>ratio</em><em> </em><em>is</em>

<em>4</em><em>7</em><em> </em><em>:</em><em> </em><em>4</em><em>3</em><em> </em><em>ratio</em>

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How many grams hydrogen in 4.5 mol H2SO4
swat32
H2SO4 ---> 2H^+ + SO4^2-

Hence n H+ = 9 mols

Mass of H = nM = (9*1) = 9g


Alternately

mass of H2SO4= nM= 4.5*98= 441

Mass of H= mass h2so4 * molar mass of H/molar mass of h2so4
Mass of H= 441 * 2/98 = 9g

4 0
3 years ago
What is the empiracle formula of amylose?
Jlenok [28]

Answer:

(C6H10O5)n

Explanation:

8 0
3 years ago
1. A part of a neuron is a(n) A. nerve B. cell body C. receptor D. acetylcholine
katovenus [111]
A neuron has 4 basic part the dendrites the cell body
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3 0
3 years ago
A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati
prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

k=\frac{2.303}{48.0}\log\frac{53500}{42520}

k=9.98\times 10^{-2}hr^{-1}

Now we have to calculate the half-life.

k=\frac{0.693}{t_{1/2}}

9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}

t_{1/2}=6.94hr

Therefore, the half life of 28-Mg in hours is, 6.94

8 0
3 years ago
YO I NEED HELP!!!! APEX PLEASE
podryga [215]
B is the correct answer for it
8 0
3 years ago
Read 2 more answers
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