Which of the following is a compound that contains 2 different elements?

Hii pls helpnme to write out the ionic equation

Oxana [17]

Answer:

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

Explanation:

According to this question, sodium carbonate reacts with sulfuric acid to form aqueous sodium sulfate, carbon dioxide and water. The balanced chemical equation is as follows:

Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)

- Next, split compounds that are aqueous into ions.

2Na+(aq) + CO32-(aq) + 2H+(aq) + SO42-(aq) → 2Na+(aq) + SO42-(aq) + CO2(g) + H2O(l)

- Next, we cancel out the spectator ions, which are ions that remain the same in the reactants and products side of a chemical reaction. The spectator ions in this equation are 2Na+(aq) and SO42-(aq).

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

- Hence, the balanced ionic equation is as follows:

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

3 0

2 years ago

"I think my bike is making a scraping noise because the bearings are not lubricated."

GuDViN [60]

Answer:

B

Explanation:

"I think my bike is making a scraping noise because the bearings are not lubricated."

4 0

2 years ago

How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100

Alona [7]

Answer:

Q = 143,921 J = 143.9 kJ.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

$Q=m*\Delta _{vap}H$

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

$Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ$

Regards!

3 0

2 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .