What is the answer to this question?
what is the answer to this question? - 1

Answers

Answer 1
Answer: I think it's 4 but I am not even 50% sure... But I think that is the answer..

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What kind of molecule can experience hydrogen bonding

Answers

A molecule with hydrogen bonded to O, N, or F.

Answer:

A molecule with hydrogen bonded to to O, N, or F

Explanation:

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How many atoms are present in 4.0 Mol of sodium

Answers

1 mole = 6.02 x 10^23 atoms (of any element)

So we are going to take our known value of 4 moles, and multiply by 6.02 x 10^23 (Avogadro's number) and we will get the number of atoms that are in 4 moles. 

(4.0 moles of Na) x (6.02 x 10^23) / (1 mole) = 2.4 x 10^24 atoms of Na
There are 2.4 x 10^24 atoms of Na in 4.0 moles


Final answer:

To find the number of atoms in 4.0 Mol of sodium, you need to multiply the number of moles by Avogadro's number. This will give you approximately 2.4088 x 10^24 atoms.

Explanation:

To calculate the number of atoms in 4.0 Mol of sodium, you need to use Avogadro's number, which is 6.022 x 1023 atoms per mole. The calculation is as follows:

Start with the number of moles given, which is 4.0 mol.Multiply this by Avogadro's number (6.022 x 1023) to get the total number of atoms:

  4.0 mol x 6.022 x 1023 atoms/mol = 2.4088 x 1024 atoms  

So, there would be approximately 2.4088 x 1024 atoms in 4.0 mol of sodium.

Learn more about Avogadro's number here:

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A second-order reaction starts with an initial concentration of 0.020 M of the reactant. If the rate constant is 1.0x10-3 M-1s-1, what is the time required for the concentration of the reactant to reach 0.010 M?

Answers

Answer:

t = 50,000s

Explanation:

Reaction is second order.

Initial conc. [A]o = 0.020 M

Rate constant = 1.0x10-3 M-1s-1

Final conc. [A] = 0.010M

Time = ?

1 / [A] = kt + 1 / [A]o

Substituting the values;

1 / 0.010 = 1.0x10-3 * t + (1/0.020)

100 - 50 = 1.0x10-3 * t

t = 50 / (1.0x10-3)

t = 50,000s

Answer:

Time required is 50000s

Explanation:

General formula of a second-order reaction is:

(1)/([A]) =(1)/([A]_0) +Kt

Where [A] is concentration of reactant after time t passed, [A]₀ is initial concentration of reactant and K is rate constant of reaction.

Replacing:

(1)/([0.010M]) =(1)/([0.020M]) +1.0x10^(-3)M^(-1)s^(-1)t

50M⁻¹ = 1.0x10⁻³M⁻¹s⁻¹ t

50000s = t

Thus, after 50000s, the reactant concentration decrease from 0.020M to 0.010M

A 1170.-gram sample of NaCl() completely reacts, producing 460. grams of Na(). What is the total mass of Cl2(g) produced?

Answers

2 NaCl --------> 2 Na  + Cl₂
  
2 mol * 23 g Na = 46 g

2*35,5 g Cl = 71 

 46 g Na --------- 71 g Cl₂
 460 g Na -------- ?

mass ( Cl₂) = 460 . 71 / 46

Mass (Cl₂) = 32660 / 46

m= 710 g of Cl₂

hope thips helps

How would you physical separate ? A.) NaCl solution
B.) sand & water
C.) Italian salad dressing

Answers


you would physically separate sand and water
The only thing you can physically separate is sand and water.

Dalton’s Law CalculationA mixture of H₂, N₂ and Ar gases is present in a steel cylinder. The total pressure within the cylinder is 675 mm Hg and the partial pressures of N₂ and Ar are, respectively, 354 mm Hg and 235 mm Hg. If CO₂ gas is added to the mixture, at constant temperature, until the total pressure reaches 842 mm Hg, what is the partial pressure, in mm Hg, of the following?A) CO₂, 167 B) N₂, 354 C) Ar, 235 D) H₂, 8629

Answers

Answer:

A) The partial presssure of CO₂ is 167 mm Hg

B) The partial presssure of N₂ is 354 mm Hg

C) The partial presssure of Ar is 235 mm Hg

D) The partial presssure of H₂ is 86 mm Hg

Explanation:

Dalton's law of partial pressures is basically expressed by the following statement:

The total pressure of a mixture is equal to the sum of the partial pressures of its components.

So initially we have:

P_(T)= total presure of the system (675 mm Hg).

P_(N_2)= partial pressure of N₂ (354 mm Hg).

P_(Ar)= partial pressure of Ar (235 mm Hg).

Using Dalton's law we can find the partial pressure of H₂:

P_(T)=P_(N_2)+P_(Ar)+P_(H_2)

675 mm Hg=354 mm Hg + 235 mm Hg + P_(H_2)

P_(H_2)= 675 mm Hg - 354 mm Hg - 235 mm Hg

P_(H_2)=86 mm Hg

If CO₂ gas is added to the mixture, at constant temperature, and the volume is the same, the difference between the new total pressure and the previous total pressure is equal to the partial pressure of CO₂.

P_(T)=P_(N_2)+P_(Ar)+P_(H_2)+P_(CO_2)

842 mm Hg= 354 mm Hg + 235 mm Hg + 86 mm Hg + P_(CO_2)

P_(CO_2)= 842 mm Hg - 354 mm Hg - 235 mm Hg - 86 mm Hg

P_(CO_2)= 167 mm Hg