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3 years ago

During which of the following chemical changes does a precipitate form

1 answer:

4 0

<span> when the cations of one reactant and the anions of a second reactant found in aqueous solutions combine to </span>form<span> an insoluble ionic solid that we call a </span>precipitate<span>.</span>

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Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to

zhuklara [117]

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

8 0

3 years ago

Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

Have a nice day!

7 0

3 years ago

What kind of bond is the result of the transfer of an electron?

serious [3.7K]

Answer:

Answer choice A

Explanation:

When an electron is transferred to another atom, both atoms involved become ions.

5 0

4 years ago

Read 2 more answers

You are given a 1.55 g mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 20 mL of water and add an e

irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.

<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>

0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

<em>Moles CaCl₂:</em>

3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>

1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture

That means mass percent of CaCl₂ is:

0.207g CaCl₂ / 1.55g * 100 =

<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>

8 0

3 years ago

PlEASE HELP! 40!

podryga [215]

Answer:

moles Na = 0.1114 g / 22.9898 g/mol=0.004846

moles Tc = 0.4562g /98.9063 g/mol=0.004612

mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g

moles O = 0.3285 g/ 15.999 g/mol=0.02053

divide by the smallest

0.02053/ 0.004612 =4.45 => O

0.004846/ 0.004612 = 1.0 => Tc

to get whole numbers multiply by 2

Na2Tc2O 9

Explanation:

Hope it right hope it helps

5 0

3 years ago