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dem82
29 days ago
7

What is the percentage by mass of solution formed by dissolving 36.0 grams of HCl in 98.0 grams of water?

Chemistry
2 answers:
Alekssandra [2.8K]29 days ago
5 0

Answer:

The mass percentage of the solution is 26.86%.

Explanation:

Hydrochloric acid mass = 36.0 g

Water mass = 98.0 g

The solution is composed of HCl and water.

Total mass of the solution =  36.0 g + 98.0 g = 134 g

Mass percentage of the solution:

\frac{\text{Mass of solute}}{\text{Mass of the solution}}\times 100

\frac{36 g}{134 g}\times 100=26.86\%

The mass percentage of the solution is 26.86%.

eduard [2.6K]29 days ago
4 0
Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.
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Alekssandra [2891]
3 first significant figure
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0.0365
6 0
1 month ago
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A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
KiRa [2857]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

Por lo tanto,

0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

3 0
1 month ago
The symbol for xenon (Xe) would be a part of the noble gas notation for the element antimony. cesium. radium. uranium.
castortr0y [2913]
Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

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A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,
Anarel [2728]

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

4 0
1 month ago
After heating the solution, you decant the liquid and wash the remaining solid with distilled water. What you are removing by th
lorasvet [2668]
What is being removed during the wash is the solvent.
7 0
21 day ago
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