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Reika [66]
4 years ago
5

What's the difference between nahco3, khco3, na2co3, and k2co3?

Chemistry
1 answer:
TEA [102]4 years ago
3 0
<span>Answer: Sodium hydrogen carbonate (NaHCO3), also called sodium bicarbonate, is baking soda. It is also important in controlling the acidity of blood. Potassium bicarbonate is a colorless, odorless, slightly basic, salty substance. Sodium carbonate Na2CO3, is the water-soluble sodium salt of carbonic acid. Potassium carbonate (K2CO3) is a white salt, soluble in water (insoluble in ethanol) which forms a strongly alkaline solution.</span>
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Give the full electron configuration for nitrogen.
garri49 [273]

Answer:

1s2 2s2 2p3

Explanation:

3 0
3 years ago
In the equation below
yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
  • The ratio between NH₃ and Cu is 2:3.
  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0
2 years ago
Which is the largest atomic radius
katrin2010 [14]

Answer:

₁₁A

Explanation:

Atomic radius

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons.

This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases.

So in given elements consider A₁₁, B₁₂, C₁₃ ans D₁₇ as sodium, magnesium, aluminium and chlorine. This is the third period and as we move form sodium to chlorine atomic radius decreases. That's why sodium has greater size.

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased

7 0
3 years ago
What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?
svp [43]
The answer is 104.9, i dont know how though
4 0
3 years ago
Read 2 more answers
The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)
valkas [14]

The production of CO_{2} is 3.3480\times 10^{4} tons/day. Converting mass into kg,

1 ton=907.185 kg, thus,

3.3480\times 10^{4} tons=3.037\times 10^{7} kg

Thus, production of CO_{2} will be 3.037\times 10^{7} kg/ day.

The specific volume of CO_{2} is 0.0120 m^{3}/kg.

Volume of CO_{2} produced per day can be calculated as:

V=Specific volume\times mass

Putting the values,

V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day

Thus, volume of CO_{2} produced per year will be:

V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year

Thus, in 4 year volume of CO_{2} produced will be:

V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}

8 0
3 years ago
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