1. How many atoms of nitrogen are there in 0.50 mol of (NH4)2CO3?

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

Answer:

Please review the following responses

Explanation:

1) A solution of 100. mL contains 19.5 g of NaCl  (3.3M)

2) 100. mL of NaCl solution at 3.00 M (3 M)

3) A solution of 150. mL holds 19.5 g of NaCl  (2.2 M)

4) The concentrations of beakers 1 and 5 are identical (1.5M)

Molar mass of NaCl = 23 + 36 = 59 g

For beaker number 3:

          59 g -------------- 1 mol

           19.5 g -------------   x

  x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For beaker number 4:

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For beaker number 5:

Molarity (M) = 0.450/0.3 = 1.5 M

Answer:- 64015 J

Solution: The calorimeter contains 4250 mL of water, which is at a temperature of 22.55 degrees Celsius.

The water's density is 1 gram per mL.

Thus, the mass of water = = 4250 grams.

After introducing the hot copper bar, the final temperature of the water reaches 26.15 degrees Celsius.

Thus, for the water = 26.15 - 22.55 = 3.60 degrees Celsius.

The specific heat capacity of water is 4.184 .

To determine the heat absorbed by the water, we can use the following formula:

where q represents heat energy, m refers to mass, and c indicates specific heat.

Now let's substitute the values into the equation to perform the calculations:

q = 64015 J

Therefore, the water absorbs 64015 J of heat.

N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).

Answer:

Endothermic: water formation from ice and a cold instant ice pack  Explanation: