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4 years ago

Si agrega un ácido a una base, ¿qué pasará?

Chemistry

2 answers:

egoroff_w [7]4 years ago

3 0

Answer:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

Explanation:

Ray Of Light [21]4 years ago

3 0

Answer:

El catión H (+) del ácido se combina con el anión OH (-) de la base para formar agua. El compuesto formado por el catión de la base y el anión del ácido se llama sal.

Explanation:

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4 years ago

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3 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

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2 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

Nezavi [6.7K]

Answer:

The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

mass = 28 mg = 0.028 g
molar mass = 90.03 g/mol
Convert mass in grams to number of moles, n:

n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol

4) Titration

Volume of base: 28.4 mL = 0.0248 liter
Concentration of base: x (unknwon)

Number of moles of acid: 2.52 mol (calculated above)

Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

5) Molarity of NaOH solution

M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

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