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Ilya [14]
3 years ago
15

Each molecule of hydrochloric acid, hcl, contains one atom of hydrogen and

Chemistry
1 answer:
galben [10]3 years ago
4 0
A.) <span>Each molecule of hydrochloric acid, HCl, contains one atom of hydrogen and "One atom of Chlorine"

Hope this helps!</span>
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help!! i tried answering these on my own and turning it into my teacher and she said they were all wrong , can somebody help me
Kobotan [32]
I don't know what your answers were but the more "stuff" there is the more energy it takes to freeze or boil it. Hope this could help
6 0
3 years ago
A compound has been shown to contain 14.5 percent hydrogen and 85.5 percent carbon by mass.
Sedaia [141]

Answer:

  • <em>Option d. Its empirical formula is CH</em><em>₂</em><em>.</em>

Explanation:

The percent composition of the compound allow you to calculate the empirical formula of the compound but is not enough to calculate either the molar mass or the molecular formula. So, since now you can discard options b. and c.

Telling that it is a hydrocarbon (option e.) is true but very vague compared with finding the empirical formula. So, you can also discard the option e.

The fact that the product has a triple bond cannot be concluded from the percent composition, you should find the molecular formula to assert whether it contains or not a triple bond. So, you could discard option a., which lets you only with choice d.

Let us find the empirical formula to be certain that it is CH₂.

1.  <u>First, assume a basis of 100 g of compound</u>:

  • H: 14.5% × 100 g = 14.0 g
  • C: 85.5% × 100 g = 85.5 g

2. <u>Divide each element by its atomic mass to find number of moles</u>:

  • H: 14.0 g / 1.008 g/mol = 14.38 mol
  • C: 85.5 g / 12.011 g/mol = 7.12 mol

3. <u>Divide both amounts by the smallest number, to find the mole ratio</u>:

  • H: 14.38 mol / 7.12 mol ≈ 2
  • C: 7.12 mol / 7.12 mol = 1.

Hence, the ratio is 2:1 and the empirical formula is CH₂.

7 0
3 years ago
Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all char
Sonbull [250]

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

 

3 0
3 years ago
By making several measurements with the same balance, a chemist obtained values of 5.224 g, 5.235 g, and 5.259 g for the mass of
stepan [7]

Answer:

a) Measurements have a good precision.

Explanation:

Accuracy is the proximity of the data to the value considered as real, in this situation we do not know the real value and we do not know if the data is accurate or not, so we can discard options b and d.

Now, precision is the proximity of the data obtained among themselves and that is what we can observe, so the appropriate answer is the option a.

6 0
3 years ago
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
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