A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution but uses a volumetric flask of vol- ume

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!

To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature. 

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L

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

The temperature difference after 3 hours is 5.16 K. Given that the moles of O₂ inhaled rate at 0.02 mole/min, which converts to 1.2 mole/hour, we know the average heat released during metabolism is 7.2 kJ/h·kg. Therefore, the amount of heat generated within 3 hours will be 7.2 kJ/h·kg multiplied by 3 hours, giving a result of 21.6 kJ/kg, or 21.6 x 10³ J/kg. Applying the formula Qp = Cp x ΔT, and taking the body's heat capacity to be 4.18 J/g·K, we find ΔT = 5.16 K.

The accurate choice is the final one.