Write the net ionic equation for the precipitation of calcium sulfide from aqueous solution

Answer:B) The Skin and The Skin cells

Explanation:

Hope this helped

Answer:

I think radiaction

Explanation:

Answer:

173.9 mL of HBr

Explanation:

We'll begin by calculating the number of mole in 6.73 g of CaCO₃. This can be obtained as follow:

Mass of CaCO₃ = 6.73 g

Molar mass of CaCO₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mole of CaCO₃ =?

Mole = mass / Molar mass

Mole of CaCO₃ = 6.73 / 100

Mole of CaCO₃ = 0.0673 mole

Next, we shall determine the number of mole of HBr that will react with 6.73 g (i.e 0.0673 mole) of CaCO₃. This can be obtained as follow:

2HBr + CaCO₃ —> CaBr₂ + H₂O + CO₂

From the balanced equation above,

2 moles of HBr reacted with 1 mole of CaCO₃.

Therefore, Xmol of HBr will react with 0.0673 mole of CaCO₃ i.e

Xmol of HBr = 2 × 0.0673

Xmol of HBr = 0.1346 mole

Thus, 0.1346 mole of HBr reacted.

Next, we shall determine the volume of HBr needed for the reaction. This can be obtained as follow:

Mole of HBr = 0.1346 mole

Molarity of HBr = 0.774 M

Volume =?

Molarity = mole / Volume

0.774 = 0.1346 / volume

Cross multiply

0.774 × volume = 0.1346

Divide both side by 0.774

Volume = 0.1346 / 0.774

Volume = 0.1739 L

Finally, we shall convert 0.1739 L to mL. This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.1739 L = 0.1739 L × 1000 mL / 1 L

0.1739 L = 173.9 mL

Thus, 173.9 mL of HBr is needed for the reaction.

Answer:

D all the above

Explanation:

Answer:

Answer:- Lions..

Explanation:

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Answer:

Yes we can calculate the solubility

The solubility of the mineral is 120 g/L

Explanation:

We have the information that the sample taken originally from the rock pool is 25.0 mL

So,

25.0 mL of the sample contained 3.000 g of the sample

10000mL of the sample now contains 1000 * 3.000/25 = 120 g

This means that 120 g of the sample dissolves in 1000mL or 1L of solution.

Therefore, the solubility of the mineral is 120 g/L

Answer:

A). Calcium chloride dihydrate (CaCl2 * 2H2O) slowly heated in a crucible to become calcium chloride (anhydrous).

Dehyration is a physical process

B). A hydrocarbon such as propane (C3H8) undergoes combustion to power a grill

Combustion is a chemical process.

C). A rock climber’s rope becomes frayed and turns the color of the rocks

This is physical process

D). A dog urinates on an air conditioner coil and the coil becomes corroded

Corrsion is a chemical process.

Explanation:

The chemical process is combustion and corrosion i.e. B and D. The physical change has been the heating of calcium chloride, and fraying of rock i.e. A and C.

A chemical process has been given as the reaction in which the composition of the sample changes. It has been an irreversible process.

A physical process has been described as a change in the physical properties of substances. It is a reversible process.

The following reaction can be given as:

A. The reaction between calcium chloride and water has been mediated by heating. It is a reversible process and does not change the chemical composition. It is a physical process.

B. The combustion results in the change in the chemical constituents of propane. It is a chemical process.

C. There has been no change in the chemical composition of materials. It has been a physical process.

D. The corrosion has resulted from the chemical change in the iron. It has been a chemical process

Thus, the chemical process is combustion and corrosion. The physical change has been the heating of calcium chloride, and fraying of rock.

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Answer:

bulb or pump, meniscus, outside

Explanation:

In order to use a pipet, place a BULB OR PUMP at the top of the pipet. Use this object to fill the pipet such that the MENISCUS of the liquid is even with the volume line. Release the liquid, touching the tip of the pipet to the side of the container if necessary to release the last drop OUTSIDE the pipet tip.

Answer:

6059.63 kcal/kg.

Explanation:

CH4 consist of Hydrogen and carbon. Therefore, Hydrogen in CH4 = 12 × 4 = 48 kg,  carbon in CH4 = 12 × 12 = 144kg.

For ethane, the amount of hydrogen present = 2 × 6 = 12kg and that of carbon in ethane = 2 × 24 = 48kg.

The weight of carbon in CO = 15 × 12 = 18kg and the weight of Hydrogen in CO = 15 × 16 = 240kg.

For hydrogen, its weight in H2 = 50 × 2 = 100kg.

For CO2, carbon has = 4 × 12 = 48 kg and oxygen has = 4 × 32 = 128kg.

For C6H6, carbon has 2 × 72 = 144kg and hydrogen has 2 × 6 = 12kg.

For N2, the amount of nitrogen= 11 × 28 = 308 kg.

For CH2= CH2, carbon has 4 × 24 = 96kg and hydrogen = 4 × 4 = 16kg.

The gross calofiric value = 1/100 [ 8080 C + 34500 + ( H - O/8) + 22405].

Where the total weight = 128 + 180+ 48 + 240 + 100 + 144 + 48 + 12 + 48 + 16 + 96 + 308 + 12 + 144 = 1524 kg.

The percentage by weight of carbon = total weight of carbon/total weight × 100.

The total Weight of carbon= 48 + 180 + 144 + 48 +144 + 96 = 660kg.

The percentage weight of carbon = 660/1524 × 100 = 43.3 %.

The percentage weight of hydrogen = total weight of hydrogen/total weight × 100.

The total weight of hydrogen = 100 + 12 + 48 + 16 +12 = 188.

The percentage weight of Hydrogen = 188/ 1524 × 100 = 12.33%.

Percentage weight of oxygen = total weight of oxygen/total weight × 100.

Percentage weight of oxygen = ( 128 + 240) / 1524 × 100 = 24.15%.

The gross calorific value = 1/100 [ 8080 × 43.3 ,+ 34500 ( 13.33 - 24.15/8) ] = 6711.02 kcal/kg.

Net calorific value = 6711.02 - 0.09 × 12.3 × 587 = 6059.63kcal/kg.

Answer:

Yes

Explanation:

Answer:

B

Explanation:

4moles of Cl2 produces 4moles of HCl therefore 12moles of Cl2 will produce 12moles of HCl

Answer:

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl

To solve this question we need to find limiting reactant. The moles of limiting reactant = Moles of NaCl produced:

Moles HCl -Molar mass: 36.46g/mol-:

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

Moles NaOH -Molar mass: 40g/mol-:

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.0045 moles * (58.44g/mol) =

Answer:

[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.

The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.

The given reaction at equilibrium state is written as:

[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]

The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]

The rate of the reaction will be r = Kb [H₂]² [S₂].

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Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

Answer:  The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93

Explanation:

Given : Volume of ethanol (solvent) = 21.10 ml

density of ethanol (solvent)= 0.789 g/ml

Mass of ethanol (solvent) = [tex]0.789g/ml\times 21.10ml=16.6g[/tex]

Mass of glycerol (solute) = 2.51 g

Mole fraction of a component is the ratio of moles of that component to the total moles present.

moles of ethanol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{16.6g}{46g/mol}=0.36mol[/tex]

moles of glycerol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{2.51g}{92g/mol}=0.027mol[/tex]

mole fraction of ethanol (solvent) = [tex]\frac{\text {moles of ethanol}}{\text {moles of ethanol + moles of glycerol}}=\frac{0.36}{0.36+0.027}=0.93[/tex]

The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93

Answer:NaOH

Explanation:all are acids except NaOH

Answer:

it is NaOH

Explanation:

Because NaOH is base, while the others are acid

Answer: The activation energy, Ea, of the reaction is 89195 Joules

Explanation:

The effect of temperature on rate constant is given by Arrhenius equation:

[tex]ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]

where  

[tex]k_2[/tex] = rate constant at temperature [tex]T_2[/tex]

[tex]k_1[/tex] = rate constant at temperature [tex]T_1[/tex]

[tex]E_a[/tex]= activation energy

R= gas constant

[tex]T[/tex] = temperature  

[tex]ln \frac{17k_1}{k_{1}} = \frac{-E_{a}}{8.314}[\frac{1}{300} - \frac{1}{278}][/tex]

[tex]2.83=\frac{-E_{a}}{8.314}[\frac{1}{300} - \frac{1}{278}][/tex]  

[tex]E_a=89195J[/tex]

The activation energy, Ea, of the reaction is 89195 Joules

The given question is incomplete. The complete question is:

The standard heat of combustion is shown in the following chemical equation [tex]C_9H_{20}(g)+14O_2(g)\rightarrow 9CO_2(g)+10H_2O[/tex][tex]\Delta H_{rxn}=-6125.21kJ/mol[/tex].  If 130 g of nonane combusts , how much heat is released?

Answer: 6211.21 kJ

Explanation:

Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.

Given :

Amount of heat released on combustion of 1 mole of nonane = 6125.21 kJ

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

1 mole of nonane [tex](C_9H_{20})[/tex] weighs = 128.2 g  

Thus we can say:  

128.2 g of nonane on combustion releases = 6125.21 kJ

Thus 130 g of [tex]C_4H_{10}[/tex] on combustion releases =[tex]\frac{6125.21}{128.2}\times 130=6211.21kJ[/tex]

Thus the heat of combustion of 130 g of nonane is 6211.21 kJ

Answer:

Independent

Explanation:

Independent Variable is the volume of the object. Dependent Variable is the mass of the object. So it

Answer:

Mg donates two electrons to O

Explanation:

Lewis dot notation uses dots and crosses to represent valence electrons on atoms.

Magnesium is a metal and would donate or lose electrons during bonding.

Oxygen is a non metal and would gain electrons during bonding.

The correct option is;

Mg donates two electrons to O

The Law of conservation of mass states the creation and destruction of mass to be impossible. The reaction between 3 moles of Potassium hydroxide and Phosphoric acid is conserved.

The law of mass of conservation states that the mass of the reactants and the products of the reaction cannot be added or removed.

In the first reaction, the mass of the atoms in the reaction is balanced, while in the second reaction the mass of potassium and hydrogen is not balanced.

In the third and fourth reactions, the mass of the elements is not balanced and not conserved hence not following the law.

Therefore, option A. [tex]\rm 3KOH + H_{3}PO_{4} \rightarrow K_{3}PO_{4} + 3H_{2}O[/tex] follows the law of conservation of mass.

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Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles  = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles  = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

The pH of the buffer after addition of 10.0 mL of 50.0mM HCl is 3.90

The Henderson-Hasselbach equation is given as:

where

The formula for calculating number of moles is:

The equation of the reaction is given below:

CH3COO- + HCl → CH3COOH + Cl-

moles of CH3COO- = initial moles - moles of HCl added

moles of CH3COOH = initial moles + moles HCl added

Moles CH3COO-

Molarity = 0.010 M

volume = 100 mL = 0.100 L

Initial moles  = 0.100 L * 0.010 M = 0.001 moles

Moles HCl = 0.010L * 0.050 M= 0.0005 moles

Moles CH3COO- = 0.001 - 0.0005 moles

Moles of CH3COO- = 0.000500 moles

Moles CH3COOH:

Molarity = 0.040 M

volume = 100 mL = 0.100 L

Initial moles  = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.004 - 0.0005 moles

Moles CH3COO- = 0.003500 moles

Substituting the calculated values:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

Therefore, the pH of the buffer after addition of the 10.0 mL 50.0mmHg HCL is 3.90

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Answer:

Br- Withdraws electrons inductively

       Donates electrons by resonance

CH2CH3 - Donates electrons by hyperconjugation

NHCH3- Withdraws electrons inductively

              Donates electrons by resonance

OCH3 -  Withdraws electrons inductively

              Donates electrons by resonance

+N(CH3)3 - Withdraws electrons inductively

Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.

Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

-NHCH3  and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.

+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.

Answer:

45 degrees at 29 so 105 is lower than 45 and 105 wants to be 32 so 45 can be higher. so then the water vapor will be 35 because 105 is too much of water gallon

If one DNA strand reads CCGTAATGCAT, the sequence of the complementary strand will be - b) GGCATTACGTA

So, DNA strand -                        C-C-G-T-A-A-T-G-C-A-T

Complementary DNA strand -  G-G-C-A-T-T-A-C-G-T-A

Thus, If one DNA strand reads CCGTAATGCAT, the sequence of the complementary strand will be - b) GGCATTACGTA

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Answer:

B is the answer

Please give brainliest

Explanation:

I play Minecraft and Dream made it once

Answer:

lol

Explanation:

lol