Cual es el volumen de la solución de hcl la cual se preparo con 3 gramos de soluto (k) al 14%

Chemistry

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

from which language this belongs to

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A point of reference is anything that seems steady that is used compare the position of an object

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Explanation:

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15.0 g of Fe and 25.0 g of sand are added to 250.0 g of water. a. Determine the percent mass of Fe, sand, and water in the mixtu

lana [24]

Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand = 8.62%

percentage mass of water = 86.205%

B. (Iron + sand + water) -------> ( iron + sand) ------> sand

C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

percentage mass of iron = 15/290 * 100% = 5.17%

percentage mass of sand = 25/290 * 100% = 8.62%

percentage mass of water = 250/290 * 100% = 86.205%

B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

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3 years ago

Determine the mass in grams of 4.69 x 1021 atoms of barium. (The

Virty [35]

Answer:

1.07 g Ba

Explanation:

Hello there!

In this case, according to the definition of the Avogadro's number and the molar mass, it is possible to say that 6.022x10^{23} atoms of barium equal one mole, and at the same time, 1 mole equals 137.327 grams of this element; thus, it is possible to say that 6.022x10^{23} atoms of barium have a mass of 137.327 grams; therefore, it i possible for us to calculate the required mass in grams as shown below:

$4.69x10^{21}atoms*\frac{137.327gBa}{6.022x10^{23} atoms} \\\\=1.07gBa$