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3 years ago

The balanced net ionic equation for precipitation of caco3 when aqueous solutions of na2co3 and cacl2 are mixed is

1 answer:

6 0

The solution for the question is the following:
NaCO3(aq) + CaCl2 (aq) --> CaCO3 + Na2Cl(aq)
Na 2+(aq) + CO3 -2 (aq) +Ca+(aq) + 2Cl-(aq) ---> CaCO3 + Na 2+(aq) + 2Cl-(aq)
Ca2+ (aq) + CO3 2- (aq) → CaCO3 (s) would be the balanced net ionic equation.

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A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling t

djverab [1.8K]

Complete Question:

A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.

Answer:

13 mol/L

Explanation:

The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:

M = n/V

The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:

n = 134/207.319

n = 0.646 mol

So, for a volume of 50 mL (0.05 L), the concentration is:

M = 0.646/0.05

M = 12.92 mol/L

Rounded to 2 significant digits, M = 13 mol/L

7 0

3 years ago

How many moles of oxygen are necessary to generate 28 moles of water, according to the following equation: 2H2+O2→2H2O

valentinak56 [21]

The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles

<h3>Balanced equation </h3>

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

<h3>How to determine the mole of oxygen needed </h3>

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

Therefore,

28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen

Thus, 14 moles of oxygen are needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

5 0

2 years ago

For the reaction shown here, 3.5 mola is mixed with 5.9 molb and 2.2 molc. what is the limiting reactant?3a+2b+c→2d

Misha Larkins [42]

<span>For equation A + 3B + 2C ---> 2D, 1 mole of A will produce 2 moles of D 3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D 2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>

7 0

3 years ago

ILL MARK BRAINLIEST :)

DanielleElmas [232]

Explanation:

According to the law of conservation of mass, mass can neither be created nor destroyed but it can simply be transformed from one form to another.

For example, $Na^{+} + Cl^{-} \rightarrow NaCl$

Mass of Na = 23 g/mol

Mass of Cl = 35.5 g/mol

Sum of mass of reactants = mass of Na + mass of Cl

= 23 + 35.5 g/mol

= 58.5 g/mol

Mass of product formed is as follows.

Mass of NaCl = mass of Na + mass of Cl

= (23 g/mol + 35.5) g/mol

= 58.5 g/mol

As mass reacted is equal to the amount of mass formed. This shows that mass is conserved.

As a result, law of conservation of mass is obeyed.

3 0

3 years ago

Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

3 years ago

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