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olga nikolaevna [1]
3 years ago
7

How to convert 35ml= dl

Chemistry
1 answer:
Nata [24]3 years ago
6 0

okay so, 35ml = 0.35 dl.

"We know (by definition) that: 1⁢ml ≈ 0.01⁢dl

We can set up a proportion to solve for the number of deciliters.

1⁢ml35⁢ml ≈ 0.01⁢dlx⁢dl

Now, we cross multiply to solve for our unknown x:

x ⁢ dl ≈ 35⁢ml1⁢ml * 0.01 ⁢ dl → x ⁢ dl ≈ 0.35000000000000003 ⁢ dl

Conclusion: 35 ⁢ ml ≈ 0.35000000000000003 ⁢ dl"

website used: https://converter.ninja/volume/milliliters-to-deciliters/35-ml-to-dl/

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A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?
olga nikolaevna [1]

 The volume that  will  be occupied at 735  torr and 57 c  is 23.12 L


 <u><em>calculation</em></u>

  • <u><em> </em></u> At STP   temperature=273 k  and  pressure=760 torr
  • <u><em> </em></u>by use of combined  gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

           T1= 273  K

           V1= 18.5 L

          P2= 735 torr

         T2=  57+273= 330 K

          V2=?

  • by making   V2 the formula of subject

     V2= T2P1V1/P2T1

       V2=  [(18.5L  x 330 k  x 760 torr)/(735 torr x 273 k)]= 23.12  L




5 0
3 years ago
What is the mass of 3.25 moles of sodium hydroxide (NaOH)?
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Answer:

<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>

Explanation:

<em>Hope this helps have a nice day :)</em>

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Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

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