An -ateor -iteat the end of a compound name usually indicates that the compound contains _____.

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:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

= 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

I hope this is useful......

Answer:

4.86×10^23 lead molecules

Explanation:

Using the provided equation, 2 moles of ammonia produces 3 moles of lead.

Thus:

2 mol NH3/ 3 mol Pb

By utilizing this proportion, we can calculate the quantities for each molecule. For 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Next, we will apply Avogadro's number to determine the total number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 lead molecules