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3 years ago

38.2mm converted into cm

Chemistry

1 answer:

Alex3 years ago

8 0

1 mm ---------- 0.1 cm
38.2 mm ------ ?

38.2 x 0.1 / 1

= 3.82 cm

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The kinetic energy of all the particles in a given sample of matter is the different. Question 11 options: True False

GalinKa [24]

Answer is true I'm voting

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3 years ago

Ethyne or acetylene can be hydrogenated to change from the triple bonded C2H2 to the double-bonded C2H4. Adding hydrogens in org

GaryK [48]

The correct answer is

C. -112.0 J/mol K

6 0

3 years ago

The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL.

yanalaym [24]

Answer:

Molarity for the solution is 1.05 mol/L

Explanation:

Mole fraction of solute = 0.0194

Solution's density = 1.0627 g/mL

We must know that sum of mole fraction = 1

Mole fraction of solute + Mole fraction of solvent = 1

0.0194 + Mole fraction of solvent = 1

Mole fraction of solvent = 1 - 0.0194 → 0.9806

Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L

With molar mass we can determine the mass of solute and solvent and then, the solution's mass

0.0194 mol . 101.1 g/ mol = 1.96 g of non electrolyte solute

0.9806 mol . 18 g/mol = 17.65 g of water

Mass of solution = mass of solute + mass of solvent

1.96 g + 17.65 g = 19.6 g (mass of solution)

Solution's density = Solution's mass / Solution's volume

1.0627 g/mL = 19.6 g / Solution's volume

Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL

Let's convert the mass from mL to L

18.4mL . 1L / 1000 mL = 0.0184 L

We have the moles of solute, so let's determine molarity

mol/L → 0.0194 mol / 0.0184 L = 1.05 M

8 0

3 years ago

Cuantos gramos de nitrato de potasio (KNO3) son necesarias para preparar 800ml de solucion al 12 % m/v

maxonik [38]

Answer:

96 g

Explanation:

Tomando en cuenta que estamos hablando de una solución de Nitrato de potasio cuya concentración es 12% m/V, debemos entonces usar la expresión de %m/V:

%m/V = msto / Vsol * 100 (1)

De esta expresión, debemos despejar la masa de soluto (msto) que es precisamente el valor que estamos buscando. Al hacer el despeje tenemos:

msto = %m/V * Vsol / 100 (2)

Ahora solo nos queda sustituir los valores dados y resolver:

msto = 12 * 800 / 100

Espero te sirva.

3 0

3 years ago

How many molecules are there in 42 grams of AIF3

Sever21 [200]

Answer:

The correct answer is - 3.012×10^23 molecules

Explanation:

The number of molecules is present in any mole of an element or compound are the same. To find the molecules you need to multiply the number of the moles of the compound by Avogadro's number:

Moles are calculated from the molecular weight, for AlF3

= 26.98 + 3×19

= 26.98 + 57

= 83.98

Then moles in 42 grams:

= 42/83.98

= 0.5001190 moles

Now the number of molecules = moles*Avogadro's number

= 0.500110×6.022×10^23

= 3.011×10^23 molecules

7 0

3 years ago