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babunello [35]
3 years ago
15

What is the PH value of CO?​

Chemistry
1 answer:
Pie3 years ago
6 0

the answer is going to be 4.5

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Rank the real gases according how they resemble an ideal gas: CO, Ne, N2, He, and NH3
Korolek [52]
<span>least to most ideal
NH3
 CO
 N2,
 Ne
He 
</span>n2, Ne, and He have dispersion forces, which are the weak intermolecular attraction for small molecules
. He is smaller than Ne, which is smaller than N2. CO and NH3 have dipole forces, which are stronger than dispersion forces. NH3 has hydrogen bonding. 

8 0
4 years ago
Why are mitochondria called the powerhouses of cells
eduard

Mitochondria are tiny organelles inside cells that are involved in releasing energy from food. This process is known as cellular respiration. It is for this reason that mitochondria are often referred to as the powerhouses of the cell.

When the breakdown products from the digestion of food find their way into the cell, a series of chemical reactions occur in the cytoplasm. This allows some of the energy locked up in these products to be released and incorporated into the universal energy supplier in cells known as ATP (adenosine triphosphate).

6 0
4 years ago
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
Calling all chemists...please help me!  Ive attached a screen shot of the problem
Olin [163]
This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.

Hope this helps!
6 0
3 years ago
Find the number of moles in 1.00 x 1023 atoms of chromium.
IRISSAK [1]

Answer:

<h2>0.17 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

n =  \frac{1.00 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{1}{6.02}  \\   = 0.166112

We have the final answer as

<h3>0.17 moles</h3>

Hope this helps you

8 0
3 years ago
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