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Serga [27]
3 years ago
7

If you have 3.01 E24 molecules of CO, how many moles of CO do you have?

Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
5 0

Answer: 5moles

Explanation:

1mole of a substance contains 6.02x10^23 molecules.

Then, 1mole of CO also contains 6.02x10^23 molecules.

If 1 mole of CO2 contains 6.02x10^23 molecules, it means Xmol of CO contains 3.01 E24 ie 3.01x10^24 molecules

Xmol of CO = 3.01x10^24 / 6.02x10^23

Xmol = 5moles

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Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution
Alla [95]

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

4 0
3 years ago
HELP!!!!!!! How many moles of AgNO3 must react to form 0.854 mol Ag?<br> mol AgNO3
iris [78.8K]
There is 0.854 moles in AgNO3
4 0
3 years ago
Read 2 more answers
Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
HELLO MAY I GET HELP AGAIN?
Novay_Z [31]

Answer:

( c ) sunlight.

Explanation:

the leaves are closer to the sunlight they require.

4 0
2 years ago
How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
AVprozaik [17]
MThe  heat  energy  required  to  raise  the  temperature   of  0.36Kg   of  copper   from   22 c   to  60  c  is  calculate  using  the  following  formula

MC delta T
m(mass)=  0.360kg  in  grams  =  0.360  x1000 = 360 g
  c(specific  heat  energy)  =  0.0920  cal/g.c
delta T =  60- 23  = 37  c

heat  energy is therefore=  360g   x0.0920 cal/g.c  x 37  c=  1225.44  cal

5 0
3 years ago
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