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3 years ago

6

(GIVING BRAINLIEST!!!)What is the correct formula for the ionic compound that forms when oppositely- charged ions of calcium and

carbonate attract?
(ANSWERS ARE IN THE IMAGE)

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

CaCO₃

Explanation:

Calcium ion is written as Ca²⁺

Carbonate ion is written as CO₃²¯

During bonding, the +2 ions from the calcium and the –2 ions from the carbonate will cancel out given a net charge of zero as shown below:

Ca²⁺ + CO₃²¯ —> CaCO₃

Thus, the correct formula for the ionic compound that forms when oppositely charged ions of calcium and carbonate attract is CaCO₃.

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How many grams of acetic acid are needed to react with 55.5 g of salicylic acid?

Julli [10]

Idk ask other person

6 0

3 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

3 years ago

The following reactions are used in batteries:Reaction I is used in fuel cells, II in the automobile lead-acid battery, and III

elixir [45]

The highest ratio is -0.613kj/g which is for automobile lead-acid battery .

Given ,

Reaction-1 used in fuel cells

Reaction-2 used in automobile lead-acid battery

Reacttion-3 used in experimental high-temperature battery for powering electric vehicles .

1) Fuel cells :

2H2(g) + O2(g) ==> 2H20 (l) , E^0cell = 1.23V

We know ,

Free energy is the maximum work done .

delta G = Wmax

=> Wmax = delta G = -4.75*10^5j/mol e^- = -4.75*10^2 kj/ mol e^-

Molecular mass of 1 mole of H2 = 2 .016 g / mol e^-

Molecular mass of 2 mole of H2 = 4.032 g/mol e^-

Molecular mass of O2 = 32 g/mol e^-

Total mass of the reactants = (4.032+32)g/mol e^- = 36.032 g/mol e^-

Ratio of Wmax to the total mass of the reactants

=Wmax/ total mass of the reactants

=( -4.75*10^2 kj/mol e^-)/36.032 g/mol e^-

=-13.2kj/g

Therefore , the ratio of Wmax to total mass of the reactants is -13.2kj/g .

2) Automobile lead-acid battery :

Pb(s) + PbO2(s) + 2H2SO4(aq) ==> 2PbSO4 (s) + 2H2O (l) , E^0cell = 2.04V

We know ,

Wmax = delta G = -3.94*10^5j/mol e^- = -3.94*10^2kj/mol e^-

molecular mass of Pb = 207.2 g /mol e^-

molecular mass of Pbo2 = 239.2g/mol e^-

molecular mass of H2SO4 = 98.008g/mol e^-

molecular mass of 2 mole of H2SO4 = 196.016g/mol e^-

total mass of the reactants = (207.2+239.2+196.016)g/mol e^- = 642.416g/mol e^-

Ratio of Wmax to total mass of the reactants

= Wmax/total mass of reactants

=(-3.94*10^2kj/mol e^-)/642.416g/mol e^-

=-0.613kj/g

Hence ,the ratio of Wmax to the total mass of the reactants of automobile lead-acid battery is -0.613kj/g .

3) Experimental high temperature battery :

2Na(l) +FeCl2(s) ==> 2NaCl (s) + Fe(s) , E^0 cell = 2.35V

We know ,

Wmax = delta G = -4.53*10^5j/mol e^- = -4.53*10^2kj/mol e^-

molecular mass of Na = 22.99 g/mol e^-

molecular mass of 2 moles of Na = 45.98g/mol e^-

molecular mass of FeCl2 = 126.85 g/mol e^-

total mass of the reactants =( 45.98+ 126.85)g/mol e^- = 172.83 g/mol e^-

Ratio of Wmax to the total mass of the reactants

= Wmax / total mass of reactants

=(-4.53*10^2kj/mol e^-)/ 172.83g/mol e^-

=-2.62kj/g

Therefore , the ratio of Wmax to the total mass of the reactants in an experimental high temperature battery is -2.62kj/g .

Hence , the highest ratio is -0.613kj/g which is of lead -acid battery .

Learn more about fuel cell here :

brainly.com/question/13603874

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4 0

2 years ago

A substance contains just copper and oxygen analysis shows that a 12.5g sample contains 11.1 of cooper and 1.4 of oxygen what is

oksian1 [2.3K]

Answer:

The formula of the compound is Cu₂O.

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 12.5 g

Mass of copper (Cu) = 11.1 g

Mass of oxygen (O) = 1.4 g

Formula of compound =?

The formula of the compound can be obtained as follow:

Cu = 11.1 g

O = 1.4 g

Divide by their molar mass

Cu = 11.1 / 63.5 = 0.175

O = 1.4 / 16 = 0.0875

Divide by the smallest

Cu = 0.175 / 0.0875 = 2

O = 0.0875 / 0.0875 = 1

Therefore, the formula of the compound is Cu₂O.

6 0

3 years ago

How many moles of H2S are produced?

Ludmilka [50]

Answer:

You can view more details on each measurement unit: molecular weight of H2S or grams This compound is also known as Hydrogen Sulfide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2S, or 34.08088 grams. Note that rounding errors may occur, so always check the results.

6 0

3 years ago