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algol [13]
3 years ago
12

How many molecules are in 237g (about a cup of water)​

Chemistry
1 answer:
ololo11 [35]3 years ago
7 0

Answer:

8.36 x 10 to the 24 power molecules of water in a cup of water

Explanation:

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How many moles are in 1.5 x 10^23 atoms of fluorine?
andrezito [222]

Answer:

0.052636002587839 moles

Explanation:

im not sure so sorry if u get it wrong...

4 0
3 years ago
One isotope of oxygen differs from another isotope of oxygen in *
Reptile [31]

Each isotope of Oxygen has a different number of neutrons

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Atomic mass is the average atomic mass of all its isotopes

Isotopes are atoms has the same number of protons but has a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Some of the isotopes of oxygen are:

\tt _8^{16}O,_8^{17}O,_8^{18}O

Each isotope has 8 protons and 8 electrons but has a different number of neutrons

For O-16: number of neutrons = 16-8 = 8

For O-17: number of neutrons = 17-8 = 9

For O-18: number of neutrons = 18-8 = 10

4 0
3 years ago
What is the acceleration of an object that goes from 3 m/s to 5 m/s in 8 s?
nikdorinn [45]

Answer: 0.25m/s2

Explanation:

Acceleration is change in velocity with time

V = final velocity = 5m/s

U =Initial velocity = 3m/s

t = time = 8s

a = Acceleration =?

a = V — U / t

a = (5 — 3) / 8

a = 2/8

a = 0.25m/s2

8 0
3 years ago
Read 2 more answers
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
O que é transferência de Calor?
irinina [24]

Answer:

I speak English can someone translate

Explanation:

7 0
3 years ago
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