The solubility of limestone, CaCO3, at 25˚C is 0.00067 g/100 mL. Write the chemical equation for the solubility equilibrium of this sparingly soluble salt in water. Then compute the molar solubility and the solubility product constant Ksp for CaCO3 at 25˚C.

Answer: The simplest and most common type is a single bond in which two atoms share two electrons. Other types include the double bond, the triple bond, one- and three-electron bonds, the three-center two-electron bond and three-center four-electron bond. ... Bonds within most organic compounds are described as covalent.

Explanation:

Answer:

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

Explanation:

Answer:

a right is 90 degrees

Answer:

QC= [O2]^3/[F2]^10

Explanation:

Answer:

The change in the internal energy of the system -878 J

Explanation:

Given;

energy lost by the system due to heat, Q = -1189 J (negative because energy was lost by the system)

Work done on the system, W = -311 J (negative because work was done on the system)

change in internal energy of the system, Δ U = ?

First law of thermodynamics states that the change in internal energy of a system (ΔU) equals the net heat transfer into the system (Q) minus the net work done by the system (W).

ΔU = Q - W

ΔU = -1189 - (-311)

ΔU = -1189 + 311

ΔU = -878 J

Therefore, the change in the internal energy of the system -878 J

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

From the question we have

mass = 14.5 × 5

We have the final answer as

Hope this helps you

Answer:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

Explanation:

We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.

Aqueous AgNO₃ and KI will dissociate in solution as follow:

AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)

KI (aq) —> K⁺(aq) + I¯(aq)

Aqueous AgNO₃ and KI will react as follow:

AgNO₃ (aq) + KI (aq) —>

Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)

Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.

Thus, the net ionic equation is:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

The net ionic equation of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex] to form [tex]AgI[/tex] precipitates is: B. [tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]

A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.

This ultimately implies that, both the charge on each atom and sum of the masses of the chemical compounds or elements in a chemical equation are properly balanced.

An ion can be defined as an atom or molecules (group of atoms) that has lost or gained one or more of its valence electrons, thereby, making it have a net positive or negative electrical charge.

First of all, we would write the dissociation equation for aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:

For [tex]AgNO_3[/tex]:

[tex]AgNO_3_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]

For [tex]KI[/tex]:

[tex]KI_{(aq)}[/tex] -----> [tex]K^{+}_{(aq)} + I^{-}_{(aq)}[/tex]

Next, we would write a chemical equation for the reaction of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:

[tex]AgNO_3_{(aq)} + KI_{(aq)}[/tex]  -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex] [tex]+ \;K^{+}_{(aq)} + I^{-}_{(aq)}[/tex] ----->[tex]AgI_{(s)} + K^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]

Note: Spectator ions refers to the ions that exist as a reactant and a product in a chemical equation because they are unchanged by the chemical reaction.

In this chemical reaction, the spectator ions are:

Finally, in order to obtain the net ionic equation, we would cancel out the two (2) spectator ions:

[tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]

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Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]

= 2.35 × 10⁻⁴ mol

Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]

Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]

Δ[O₂]  = 0.04626 M

The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

= [tex]\dfrac{0.04626}{60}[/tex]

= 7.71 × 10⁻⁴ M/s

Answer:

2 and a half hours

Explanation:

Time is equal to distance over speed

Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

Answer:evaporation

Explanation:

Answer:

Pretty sure its evaporation, if its not I'm very sorry.

Answer:

The answer will be 29.4˚C.

Explanation:

Using the formula 5(˚F-32)/9, plug in the numbers and you'll get 29.4˚C.

Answer:

D represents the element nitrogen which will gain three valence electrons forming a 3 ion.

Answer:

answer is

Explanation:

D

Answer:

red bricks

Explanation:

right on edg 2020

Red bricks should you use so that the area is cool in terms of temperture. Therefore, option B is correct.

Red bricks can be used in the construction of structures such as buildings, foundations, arches, pavement, and bridges. These can also be used for aesthetic purposes such as landscaping, face work, and a variety of other architectural purposes.

Traditional red bricks are saw to be more robust, and structures constructed with them are stronger than hollow block structures.Red brick can be warm, with rust or terracotta undertones. They could also be cool and more burgundy.

Thus, Red bricks should you use so that the area is cool in terms of temperture, option B is correct.

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#SPJ6

Answer:

Problem

Explanation:

The given statement is a problem. It states the problem that the house of the speaker has been undergoing with. This problem gives rise to the Hypothesis in which the 'why' question is asked. The reason of the crumbling of the wood is stated in the hypothesis. Any reason placed of the happening of the event is stated to be hypothesis.

Answer:

acidic because of electrical issues and the body of electrical equipment

Answer:

Compaction.

Explanation:

All the others are before this.

Answer:

b. gelatin

Explanation:

a homogeneous noncrystalline consisting of large molecules or ultramicroscopic particles of one substance.

Answer:

Nervous and Excretory

Explanation:

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

Nervous and Excretory systems are working together in this scenario.

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

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Answer:

height = 4.74 cm

density = 2.23 g/ cm³

Explanation:

Mass of metal = 0.650 kg (650 g)

Width = 0.136 m    

Length = 0.0451 m  

Volume of metal = 291 cm³

Height in cm = ?

density of metal =?

Solution:

Width = 0.136 m    (0.136 m×100 cm/1m = 13.6 cm)

Length = 0.0451 m  (0.0451 m×100 cm/1m = 4.51 cm)

First of all we will calculate the height:

Volume = height× width× length

291 cm³ =      h     × 13.6 cm × 4.51 cm

291 cm³ =      h     × 61.34 cm²

h = 291 cm³ / 61.34 cm²

h = 4.74 cm

Density:

d = m/v

d = 650 g/291 cm³

d = 2.23 g/ cm³

Answer:

i

[tex]J_{m} = 20 [/tex]

ii

[tex]J_{m} = 22.5 [/tex]

Explanation:

From the question we are told that

  The first temperatures is [tex]T_1 =  25^oC =  25 +273 =298 \ K[/tex]

   The second temperature is  [tex]T_2 =  100^oC =  100 +273 = 373 \ K[/tex]

Generally the equation for  the most highly populated rotational energy level is mathematically represented as

     [tex]J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}[/tex]

Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]

Also  

      B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]

   Generally the energy require per mole to move 1 cm is  12 J /mole

So   [tex]0.244 \ cm^{-1}[/tex]  will require x J/mole

           [tex]x =  0.244 *  12[/tex]

=>          [tex]x =  2.928 \ J/mol [/tex]

So at the first temperature

     [tex]J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 20 [/tex]

So at the second temperature

           [tex]J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 22.5 [/tex]

Answer:

Explanation:

We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.

From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.

Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.

Answer:

a) - 0.2 M

b) - 0.2 M

c)- 0

Explanation:

The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:

MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol

a). Molarity = moles CuBr₂/1 L solution

moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol

Volume in L = 375 mL x 1 L/1000 mL = 0.375 L

M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M

b). When is added to water, CuBr₂ dissociates into ions as follows:

CuBr₂ ⇒ Cu²⁺ + 2 Br⁻

We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:

0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M

c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).

Answer:

C) an increase in rate of reaction because reactant molecules collide with greater energy

Explanation:

Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.

The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.

Answer:

Explanation:

We shall use the formula S₁V₁ = S₂V₂

S₁ = .1180 M , V₁ = 25.98 mL

S₂ = ? , V₂ = 52.50 mL

.1180 M  x 25.98 = 52.50 x S₂

S₂ = .0584 M

Molarity of the acid solution = .0584 M .

The concentration of the acid solution is 0.058 M.

The equation of the reaction is;

CH3COOH(aq) + KOH(aq) ----->  CH3COOK(aq) + H2O(l)

The following are known;

Concentration of base CB = 0.1180 M

Volume of base VB =  25.98 mL

Concentration of acid CA = ?

Volume of acid VA = 52.50 mL

Number of moles of acid NA = 1

Number of moles of base NB = 1

Using;

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA=  CBVBNA/VANB

CA = 0.1180 M × 25.98 mL × 1/52.50 mL × 1

CA = 0.058 M

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Answer : [tex]BaCl_2[/tex] reagent predict to be in excess.

Explanation : Given,

Concentration of [tex]BaCl_2[/tex] = 0.10 M

Volume of [tex]BaCl_2[/tex] = 7.50 mL = 0.0075 L       (1 L = 1000 mL)

Concentration of [tex]KIO_3[/tex] = 0.10 M

Volume of [tex]KIO_3[/tex] = 7.50 mL = 0.0075 L

First we have to calculate the moles of [tex]BaCl_2[/tex]  and [tex]KIO_3[/tex].

[tex]\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2\times \text{Volume of }BaCl_2[/tex]

[tex]\text{Moles of }BaCl_2=0.10M\times 0.0075L=0.00075mol[/tex]

and,

[tex]\text{Moles of }KIO_3=\text{Concentration of }KIO_3\times \text{Volume of }KIO_3[/tex]

[tex]\text{Moles of }KIO_3=0.10M\times 0.0075L=0.00075mol[/tex]

Now we have to calculate the excess and limiting reagent.

The balanced equilibrium reaction will be:

[tex]BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl [/tex]

From the balanced reaction we conclude that

As, 2 mole of [tex]KIO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]

So, 0.00075 moles of [tex]KIO_3[/tex] react with [tex]\frac{0.00075}{2}=0.000375[/tex] moles of [tex]BaCl_2[/tex]

From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KIO_3[/tex] is a limiting reagent and it limits the formation of product.

Hence, [tex]BaCl_2[/tex] reagent predict to be in excess.

Answer:

D

Explanation:

It is a solid everything that is a solid is heavier