How many milliliters of an aqueous solution of 0.227 M sodium carbonate is needed to obtain 3.55 grams of the salt?

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

Answer:

7.5 x 10^14 Hz

Explanation:

Answer:

Electronegativity  = 1.87.

Ionic radius = 109 pm.

Atomic radius = -39 pm

First ionization energy = 410 kJ/mol

Explanation:

Hello!

In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.

In such a way, the required comparison is written below:

Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.

Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.

Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm

First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol

It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.

Best regards!

Oxygen and Beryllium are the elements that pertain to the same period and different groups 6A and 2A.

Ionic radius, atomic radius, electronegativity and first ionization energy are periodic qualities that have selective trends.

Differences in the trends are:

This can be explained as:

= 140 pm (oxygen)- 31 pm (beryllium)

= 109.

= 73 pm (oxygen) - 112 pm (beryllium)

= -39 pm

= 3.44 (oxygen) - 1.57 (beryllium)

= 1.87

= 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium)

= 410 kJ/mol

Therefore, these data tell that ionic radius, electronegativity and first ionization energy increases from left to right in a period whereas, the atomic radius increases from right to left in a period.

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Answer:

A test variable is a user-defined, name-value pair that stores and refers to information throughout a test and between tests. For example, in a study looking at how tutoring impacts test scores, the dependent variable would be the participants' test scores, since that is what is being measured...

The control group is defined as the group in an experiment or study that does not receive treatment by the researchers and is then used as a benchmark to measure how the other tested subjects do. For example, The control group is used in an experiment as a way to ensure that your experiment actually works. Your experimental group would be given the Gatorade and your control group would be given regular water. The conditions must be exactly the same for all members of the experiment.

A limiting factor is anything that constrains a population's size and slows or stops it from growing. Some examples of limiting factors are biotic, like food, mates, and competition with other organisms for resources. Limiting factors are usually expressed as a lack of a particular resource.

Infectious agents commonly spread through the direct transfer of bacteria, viruses, or other germs from one person to another. This can happen when an individual with the bacterium or virus touches, kisses, or coughs or sneezes on someone who isn't infected.

Explanation:

The number of moles of  Al₂ O₃ : 0.2

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

2Al(s) + 3FeO(s) ⇒ 3Fe(s) + Al₂ O₃(s)

0.6 mol of Fe = 0.6 mol of FeO

mol of Al₂ O₃ :  mol of FeO = 1 : 3

[tex]\tt \dfrac{1}{3}\times 0.6=0.2[/tex]

Answer:

Chemical Equation: (NH4)2S(aq) + 2KOH(aq) -> K2S(aq) + 2NH4OH(s)

Complete Ionic Equation: 2NH4^+(aq) + S^2-(aq) + 2K^+(aq) 2OH^-(aq) -> 2K^+(aq) + S^2- (aq) + 2NH4OH(s) 2NH4^+(aq) + OH^-(aq) -> NH4OH(s)

Explanation:

Ammonium sulfate and potassium hydroxide interact in this process to produce aqueous potassium sulfate, gaseous ammonia, and liquid water as byproducts. In essence, an acid-base reaction is taking place here, with the ammonium ion contributing a hydrogen ion to the potassium hydroxide's hydroxide ion.

Chemical formulae and symbols are used to express a chemical reaction in a chemical equation.

A chemical equation represents a chemical reaction using symbols and the chemical formulas of the reactants and products involved. In this process, ammonia and hydrochloric acid are reacting to generate ammonium chloride. For solids, the symbol is (s).

( NH₄ ) 2S ₍ aq ₎ + 2KOH ₍ aq ₎ ⇒ K₂S ₍ aq ₎  + 2NH₄

Thus, Aqueous solutions of ammonium sulfide and potassium hydroxide are mixed to formed potassium sulphide and ammonia.

To learn more about chemical equation follow the link below;

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Answer:

B- heat from an oven

Explanation:

temperature isn’t energy

Answer:

NaBr + Cl2 = NaCl + Br2 - Chemical Equation Balancer.

Explanation:

Answer:

The major jet streams on Earth are westerly winds that flow west to east.

Answer:

565 kJ

Explanation:

[tex]H_v[/tex] = Heat of vaporization of water at 100°C = 2260 J/g

m = Mass of water = 250 g

Heat required is given by

[tex]Q=mH_v[/tex]

[tex]\Rightarrow Q=250\times 2260[/tex]

[tex]\Rightarrow Q=565000\ \text{J}[/tex]

[tex]\Rightarrow Q=565\times 10^3\ \text{J}[/tex]

[tex]\Rightarrow Q=565\ \text{kJ}[/tex]

The heat required to vaporize the water is 565 kJ.

Answer:

True

Explanation:

Answer:

In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.

Answer:

d. IF3

Explanation:

The Octet rule posits that atoms gain, atom lose, or share electrons in order to have a full valence shell of 8 electrons. This statement occurs when atoms also combine to form molecules until they attain or share eight valence electrons either by losing or gaining eletrons.

From the given options, a valid Lewis structure that cannot be drawn without violating the octet rule is IF3

Heat required : 1.523 kJ

Reaction

N₂+2O₂⇒2NO₂ AH = 67.7 kJ

mol of N₂:(use Pv=nRT)

[tex]\tt n=\dfrac{3.32\times 0.304}{0.082\times 373}=0.033[/tex]

mol of O₂:

[tex]\tt n=\dfrac{3.32\times 0.41}{0.082\times 373}=0.045[/tex]

Limiting reactant :

N₂ : O₂

[tex]\tt \dfrac{0.033}{1}\div \dfrac{0.045}{2}=0.033:0.0225[/tex]

Limiting reactant : O₂(smaller ratio)

mol NO₂ = mol O₂ = 0.045

2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :

[tex]\tt \dfrac{0.045}{2}\times 67.7=1.523~kJ[/tex]

The heat required to produce the maximum yield of NO2 is 1.52 kJ.

The balanced reaction equation is;

N2(g) + 2 O2(g) -> 2 NO2(g)

The number of moles of each gas is obtained from the ideal gas equation as follows;

For N2;

P = 3.32 atm

V = 304 ml or 0.304 L

T = 100ºC + 273 = 373 K

n = ?

R = 0.082 atmLK-1mol-1

From;

PV = nRT

n = PV/RT

n = 3.32 atm × 0.304 L/0.082 atmLK-1mol-1 × 373 K

n = 0.033 moles

For O2;

PV = nRT

n =  PV/RT

V = 410 ml or 0.410 L

P = 3.32 atm

R = 0.082 atmLK-1mol-1

T = 100°C + 273 = 373 K

n =  3.32 atm × 0.410 L/0.082 atmLK-1mol-1 × 373 K

n = 0.045 moles

1 mole of N2 reacts with 2 moles of O2

0.033 moles of N2 reacts with 0.033 moles × 2 moles/1 mole

= 0.066 moles

2 moles of O2 yields 2 moles of NO2

0.045 moles O2 yields 0.045 moles × 2 moles/2 moles = 0.045 moles

If 2 mole of NO2 requires 67.7 kJ

0.045 moles of NO2 requires 0.045 moles  × 67.7 kJ/2 mole

= 1.52 kJ

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Explanation:

The number of orbitals in the sublevels are given below:

     Sublevels                   Orbitals

        s                                  1

        p                                 3

        d                                 5

        f                                   7

a. ls       -  1 orbital

b. 5s      - 1 orbital

c. 4d     - 5 orbitals

d. 4f       - 7 orbitals

e. 7s       - 1 orbital

f. 3p       - 3 orbitals

g. Entire 5th principal energy level : for s, p ,  f  

          1 + 3 + 5 + 7 + 9  = 25

h. 6d   - 5

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

Answer:

In this case, 1 mole of liquid oxygen requires 3.41 kJ of heat in order to perform that liquid → solid. your sample from grams to moles by using oxygen's molar mass.

Answer:

b

Explanation:

b

Answer:

Organelles allow for various functions to occur in the cell at the same time. Despite their fundamental similarities, there are some striking differences between animal and plant cells (see Figure 1). Animal cells have centrosomes (or a pair of centrioles), and lysosomes, whereas plant cells do not.

Explanation:

Answer:

you are correct, the answer is 2 g

Answer:

sulfuric acid ..............

Answer:

2LiCl + 3O2  

Explanation:

Answer:

(C) 2LiCl + 3O2

Explanation:

sub to pewds

Answer:

A.

............

Explanation:

Ductility

Answer:

2 moles.

Explanation:

2NaOH + H2SO4 -->  Na2SO4 + 2H2O

This shows that 2 moles of NaOH give 1 mole of Na2SO4.

So 4 moles produces 2 moles of Na2SO4.

Answer:

I think it would be b

Explanation:

Answer:

Explanation:

The acceleration of gravity of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

f is the force

m is the mass

From the question we have

[tex]a = \frac{800}{75} = \frac{32}{3} \\ = 10.666666...[/tex]

We have the final answer as

Hope this helps you

Answer:

F2

Explanation:

F2 is the only element with only 1 element and eveything else is formed by 2 non-metals

Answer:d I believe

Explanation:

The specific heat of metal = 0.897 J/g° C  

Heat absorbed can be calculated using the formula:  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q = 8.793 J

m = 2 g

∆t = 4.9 °C

[tex]\tt c=\dfrac{Q}{m.\Delta t}\\\\c=\dfrac{8.793}{2\times 4.9}\\\\c=0.897~J/g^oC[/tex]

Answer:

5.3 mol CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Explanation:

Step 1: Define

3.2 × 10²⁴ molecules CO₂

Step 2: Convert

[tex]3.2 \cdot 10^{24} \ mc \ CO_2(\frac{1 \ mol \ CO_2}{6.022 \cdot 10^{23} \ mc \ CO_2} )[/tex] = 5.31385 mol CO₂

Step 3: Check

We are given 2 sig figs. Follow sig fig rules.

5.31385 mol CO₂ ≈ 5.3 mol CO₂