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Ipatiy [6.2K]
2 years ago
11

how much would the boiling point of water increase if 4 mol of NaCl were added to 1 kg of water (Kb=0.51 C/(mol/Kg) for water an

d i = 2 for NaCl
Chemistry
2 answers:
Bad White [126]2 years ago
6 0

Answer:

4.08

Explanation:

Vesnalui [34]2 years ago
5 0

Answer:

4.08°C

Explanation:

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Radioactive decay is likely to occur when ... radioactive decay is likely to occur when ... protons break into neutrons and elec
xenn [34]
Radioactivity or radioactive decay is the process that occurs when unstable isotopes or atoms release energy by emitting radiations such as ,gamma radiations, alpha radiations and beta radiations to attain stability. Therefore, in this case decay is likely to occur when a given atom has two many neutrons in its nucleus in order to attain stability.
3 0
3 years ago
Read 2 more answers
Why are valence electrons important when considering the bonds that form between atoms g
salantis [7]

Answer:Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.

Explanation::)

7 0
3 years ago
n-Butane fuel (C4H10) is burned with the stoichiometric amount of air. Determine the mass fraction of each product. Also, calcul
tia_tia [17]

Answer:

  1. 0.1852
  2. 0.0947
  3. 0.7201
  4. 3.0345 kg CO_{2} / Kg C_{4} H_{10}
  5. 15.3848 Kg air / kg C_{4} H_{10}

Explanation:

Molar masses of each product are :

Butane = 58 kg /kmol

Oxygen = 32 kg/kmol

Nitrogen = 28 kg/kmol

water = 18 kg/kmol

<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>

= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )

= 176 / 950.32

= 0.1852

<em><u>2) Calculate the mass fraction of water </u></em>

= ( 5 * 18 ) /  (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))

= 90 / 950.32

= 0.0947

<em><u>3) Calculate the mass fraction of Nitrogen </u></em>

= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))

= 684.32 / 950.32

= 0.7201

<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>

Mco2 = ( 4 * 44 ) / 58  = 3.0345 kg CO_{2} / Kg C_{4} H_{10}

<u>5) Mass of Air required per unit of fuel mass burned </u>

Mair = ( 6.5 * 32 + 24.44 *28 ) / 58  = 15.3848 Kg air / kg C_{4} H_{10}

5 0
3 years ago
Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 42. g of butane is m
taurus [48]

Answer:

127 grams of carbon dioxide

Explanation:

We need to determine the chemical equation first. Butane has a chemical formula of C_4H_{10}, oxygen is O_2, carbon dioxide is CO_2, and water is H_2O. The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:

C_4H_{10}+O_2 ⇒ CO_2+H_2O

But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:

C_4H_{10}+O_2 ⇒ 4CO_2+5H_2O

However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:

C_4H_{10}+6.5O_2 ⇒ 4CO_2+5H_2O

Finally, multiply everything by 2 to get whole number coefficients:

2C_4H_{10}+13O_2 ⇒ 8CO_2+10H_2O

Ah, now we can actually get to the problem!

We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.

The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:

42 gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}} *\frac{8molCO_2}{2molC_4H_{10}} =2.8896molCO_2

150 gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{13molO_2} =2.8846molCO_2

Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.

Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:

2.8846molCO_2*\frac{44.01gCO_2}{1molCO_2} =127gCO_2

Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.

<em>~ an aesthetics lover</em>

4 0
3 years ago
Durante una determinación de acidez en una muestra contaminada, agregas como indicador fenolftaleina y torno a violeta. ¿Qué sus
algol13

Answer:

HCl, ya que la sustancia es una base que se debe titular con un ácido.

Explanation:

¡Hola!

En este caso, teniendo en cuenta la descripción inicial de la sustancia, la cual se torna violeta cuando se le agrega la fenolftaleína, es posible inderir que esta sustancia es una base con pH básico. Ahora bien, en torno a la especificación de un proceso de titulación, es claro que dicha base debe ser titulada con un ácido, y este caso, con HCl, ácido chlorhídrico, con el fin the alcanzar el punto de equivalencia.

¡Saludos!

6 0
3 years ago
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