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3 years ago

¿Todos los solutos se disuelven en cualquier solvente?

Chemistry

1 answer:

Katen [24]3 years ago

6 0

Answer:

Explanation: No. Todos los solutos tienen una solubilidad diferente, lo que significa que todos tienen una cantidad única de gramos que se disolverían en el mismo solvente. una solución que ha disuelto todo el soluto que puede contener a una temperatura determinada.

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Electrons determine the of an atom

g100num [7]

Answer:

if u know the number of electrons u also know the number of protons present in the atom.

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2 years ago

Given the following information:

bonufazy [111]

Answer:

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2 years ago

PLSSS HELP MEE 100 POINTS!!!!!!! How many atoms of each element does magnesium hydroxide, Mg(OH)2, contain? Magnesium = 1 atom,

Basile [38]

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3 years ago

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The law of conservation of matter states that matter cannot be created or destroyed. This law applies when a chemical change occ

lora16 [44]

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2 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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1 year ago