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prohojiy [21]
3 years ago
5

MY NOTES ASK YOUR TEACHER Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of N

aOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H3O2. (a) What is the stoichiometry of HC2H3O2 to Ca(OH)2
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer:

Two moles of HC2H3O2 react with one mole of Ca(OH)2 to produce one mole of calcium acetate and two moles of water.

Explanation:

HC2H3O2 is Acetic acid that can also be represented as (CH3COOH).

when Ca(OH)2 reacts with Acetic acid the product formed will be Calcium acetate and water

Chemically the reaction can be represented as

2CH 3 COOH + Ca(OH) 2    →      Ca(CH 3 COO) 2   +    2H 2 O

Two moles of CH3COOH react with one mole of Ca(OH)2 to produce one mole of Ca(CH3COO)2 and two moles of H2O.

You might be interested in
a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


4 0
3 years ago
AlCl3 + NaOH = NaClO2 + Al(OH)3<br> balance the equation!<br> *dont forget the O2 in NaClO2!!
Verdich [7]

Answer:

AlCl₃  + 3NaOH  → 3NaCl  +   Al(OH)₃

Explanation:

Problem is to balance the given chemical equation:

      Equation:  

              AlCl₃  + NaOH  → NaClO₂   +   Al(OH)₃

Balancing a chemical equation involves the conservation of atoms on both sides of the equation.

We can use a very simple mathematical method to balance the above equation;

           aAlCl₃  + bNaOH  → cNaCl   +   dAl(OH)₃

a, b, c and d are coefficients that will balance the equation:

  Conserving Al:  a  = d

                       Cl:   3a = c

                       Na:   b = c

                       O:      b = 3d

                       H:      b = 3d

Assume that a = 1;

                       c = 3

                       b = 3

                      d = 1

            AlCl₃  + 3NaOH  → 3NaCl  +   Al(OH)₃

O₂ in from of the NaClO₂ flouts the rule of chemical combination

6 0
3 years ago
Explain how creativity can play a role in the construction of scientific questions and hypotheses.
ozzi
Like if you imagine making something you create a hypothesis about was to make that happen
6 0
4 years ago
El osmio es un metal sólido que tiene una densidad
pentagon [3]

Respuesta:

4.42 × 10⁻⁶ m³

Explicación:

Paso 1: Información provista

  • Densidad del osmio (ρ): 22600 kg/m³
  • Masa de osmio (m): 100 g

Paso 2: Convertir 100 g a kilogramos

Usaremos el factor de conversion 1 kg = 1000 g.

100 g × 1 kg/1000 g = 0.100 kg

Paso 3: Calcular el volumen ocupado por 0.100 kg de osmio

La densidad es una propiedad intensiva, igual al cociente entre la masa y el volumen.

ρ = m/V

V = m/ρ

V = 0.100 kg / (22600 kg/m³) = 4.42 × 10⁻⁶ m³

5 0
3 years ago
Please help me I'm really confused on how to do this. Can someone break it down for me? (stoichiometry)
bazaltina [42]

Answer:

so with every stoichiometry problem with a mass it will make it so you can do the conversion factor with reactants or products.

if you dont understand unit conversions try to study how to set it up. anyways

a.) C12H22O11 has a mass of 342.01 Grams per mole

divide 1.202 G by 342.01 G to get 0.004 miles

b.) you're just taking the AMU of each element in the chemical multiply it by how many there is of it in the chemical, then divide it by the mass of a mole of the chemical.

c.) you take your answers of part b and multiply them by Avogadro's number

6 0
4 years ago
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