an element has 8 protons and 8 neutrons in its nucleus and 8 electrons revolving around the nucleus. What is the atomic number and atomic mass of this element?

Answer: 38 chlorine atoms are in 3 molecules of HCI

Explanation:

Final answer:

In 2 molecules of ICl, there would be 2 atoms of Chlorine, as the subscript of Chlorine in the compound ICl is 1.

Explanation:

In a compound, the subscript of an element indicates the number of atoms of that element present per molecule of the compound. For the ICl compound, there is one atom of Chlorine in one molecule. Therefore, in 2 molecules of ICl, there would be 2*1 = 2 atoms of Chlorine.

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Answer:

1 litre of 1.0 M NaCl

Explanation:

When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;

NaCl(s) ------> Na^+(aq) + Cl^-(aq)

Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.

Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl

Final answer:

The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.

Explanation:

Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.

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3,5, and 1 i think maybe i uhm yeah

Answer:

It produces water.

Explanation:

H+   +    OH-    produces    H2O.

It is a type of Neutralization reaction.

The heat that is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling

C) 157 kJ

Heat

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

Q=m* c* delta T

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat = 500 x 4.184 x 75

Heat = 156900 J

Heat = 156.9 KJ

Heat ~ 157.0 KJ (3.D.P)

Thus, the correct answer is C.

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Answer:

C

Explanation:

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat needed= 500 x 4.184 x 75

= 156900 J

= 156.9 KJ

~ 157.0 KJ (3.D.P)

The number of moles of HBr and the concentration of the Ca(OH)2 solution is:

The number of moles HBr is = 0.001825

The concentration of Ca(OH)2 is= 0.009125 M

What is the Acid solution for neutralization?

Data given as per question:

The Volume of the Ca(OH)2 is = 100.0 mL = 0.100 L

Then, Molarity of HBr is = 5.00 * 10^-2 M

After that Volume of HBR is = 36.5 mL = 0.0365 L

When The balanced equation is:

Then, Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Then the Calculate molarity of Ca(OH) 2

After that b*Va* Ca is = a * Vb*Cb

Then ⇒with b = the coefficient of HBr is = 2

Now, ⇒with Va = the volume of Ca(OH)2 is = 0.100 L

After that ⇒with ca is = the concentration of Ca(OH)2 = TO BE DETERMINED

Now, ⇒with a = the coefficient of Ca(OH)2 = 1

Then ⇒with Vb is = the volume of HBr = 0.0365 L

Now, ⇒with Cb is = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

Then 2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Now, Ca is = (0.0365*0.05) / 0.200

Therefore, Ca is = 0.009125 M

After that, we Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

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Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles