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2 years ago

How many elements are in C5H4O2S

Chemistry

1 answer:

elena-s [515]2 years ago

8 0

Answer:

4 elements

Explanation:

The four elements would be:

C-Carbon

H-Hydrogen

O-Oxygen

S-Sulfur

Hope this helps :)

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Iontophoresis is a noninvasive process that transports drugs through the skin without needles. In the photo, the red electrode i

anyanavicka [17]

Answer:

Follows are the explanation to this question:

Explanation:

When the drug is negatively charged, its negative electrolyte is annihilated to just the positive electrode. It is enticed, and it may not have a picture showing the electrode, however, We suppose that electrodes from either side of a skin slice. Its negative electrode will bypass or push thru the skin if in front of the counter terminal this becomes a red-positive electrode.

8 0

2 years ago

5 atoms Pb to mol Pb

jek_recluse [69]

Answer:

0,83 x 10-²³ mol

Explanation:

n=N/Na

n=5:(6,02 x 10²³)

5 0

3 years ago

The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati

Pepsi [2]

Answer: 3.72 M

Explanation:

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

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

where,

k = rate constant = $3.66\times 10^{-3}s^{-1}$

t = age of sample = 15.0 minutes

a = let initial amount of the reactant = 10.0 M

a - x = amount left after decay process = ?

$15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}$

$\log\frac{100}{(a-x)}=1.43$

$\frac{100}{(a-x)}=26.9$

$(a-x)=3.72M$

The concentration of $H_2O_2$ in a solution after 15.0 minutes have passed is 3.72 M

7 0

3 years ago

The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse

Umnica [9.8K]

Answer : 121.5 μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

 $\frac{N}{ N_{0} } = e^{-( \frac{0.693 X T_{2} }{T_{1}})$

3.8 / $N_{0}$ = $e^$ ((0.693 X 162.5 ) / 32.5) = 121.5

On solving we get, The initial activity as 121.5 μci

5 0

2 years ago

Read 2 more answers

A gold ring with a mass of 16g was dropped in the snow and its temperature dropped from 35°C to 0°C. How much heat was released

Luda [366]

Answer:

-72.8 joules

Explanation:

just finished the test

4 0

2 years ago