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3 years ago

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g When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solut

ion of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)

1 answer:

Elden [556K]3 years ago

5 0

Answer:

When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solution of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction.

Explanation:

The word equation for the reaction is:

Copper (II) chloride(aq) + sodium carbonate (aq) ->sodium chloride (aq) + copper carbonate(s)

The balanced chemical equation of the reaction is:

$CuCl_2(aq)+Na_2CO_3(aq)->2NaCl(aq)+CuCO_3(s)$

The complete ionic equation is:

$Cu^2+(aq)+2Cl^-(aq)+2Na^+(aq)+CO_3^2^-(aq)->2Cl^-(aq)+2Na^+(aq)+CuCO_3(s)\\$

The net ionic equation is obtained from the complete ionic equation after removing the spectator ions:

$Cu^2^+(aq)+CO_3^2^-(aq)->CuCO_3(s)$

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If you have a graduated cylinder containing 15.50 mL of water and this volume changes to 17.97 mL after a metal with a mass of 1

ollegr [7]

Answer:

Explanation:

density = mass / volume

mass = 17.95 grams

volume = 17.97 - 15.50 mL = 2.45 mL = 2.45 cc

density = 17.95 /(2.45) = 7.327 grams / cc

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3 years ago

a 1.642 g sample of calcium bromide is dissolved in enough water to give 469.1 mL of solution what is the bromide ion concentrat

aleksklad [387]

The bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.

<h3>How to calculate concentration?</h3>

The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.

No of moles of calcium bromide is calculated as follows:

moles = 1.642 ÷ 199.89 = 8.215 × 10-³moles

Molarity = 8.215 × 10-³moles ÷ 469.1mL = 1.75 × 10-⁵M

Therefore, the bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.

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2 years ago

Chlorine has two natural isotopes, chlorine-35 and chlorine-37.

Alex Ar [27]

Answer:

The relative atomic mass of chlorine depends on the ratio between the abundance of these two naturally-occurring isotopes.

Explanation:

The relative atomic mass of an element is a weighted average of the atomic mass of its naturally-occurring isotopes. The relative abundance of each isotope gives its weight in this weighted average.

For these two naturally-occurring isotopes of chlorine:

The relative atomic mass of $^{35}{\rm Cl}$ is approximately $34.969$ daltons. The relative abundance of this isotope in nature is approximately $0.758$ .
The relative atomic mass $^{37}{\rm Cl}$ is approximately $36.966$ daltons. The relative abundance of this isotope in nature is approximately $0.242$ .

$\begin{array}{|c|c|c|}\cline{1-3} \text{Isotope} & \text{Atomic Mass} & \text{Relative Abundance}\\ \cline{1-3} ^{35}{\rm Cl} & \approx 34.968\; \rm Da} & \approx 0.758 \\ \cline{1-3} ^{37}{\rm Cl} & \approx 36.966\; \rm Da & \approx 0.242 \\ \cline{1-3}\end{array}$ .

$\begin{aligned}&\text{relative atomic mass of Cl} \\ &= \text{atomic mass of $^{35}{\rm Cl}$} \times \text{relative abundance of $^{35}{\rm Cl}$} \\&\quad + \text{atomic mass of $^{37}{\rm Cl}$} \times \text{relative abundance of $^{37}{\rm Cl}$} \\ &\approx 34.968\; \rm Da \times 0.758 + 36.966\; \rm Da \times 0.242 \\ &\approx 35.45\; \rm Da \end{aligned}$ .

The relative abundance of $^{35}{\rm Cl}$ is much higher than that of $^{37}{\rm Cl}$ . Consequently, the relative atomic mass of the element $\rm Cl$ is closer to the atomic mass of $^{35}{\rm Cl}\!$ than that of $^{37}{\rm Cl}\!$ .

7 0

3 years ago

A 6.35 l sample of carbon monoxide is collected at 55.0◦c and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦c?

Sonja [21]

Answer : The final volume of gas will be, 5.46 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.892 atm

$P_2$ = final pressure of gas = 1.05 atm

$V_1$ = initial volume of gas = 6.35 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $55.0^oC=273+55.0=328K$

$T_2$ = final temperature of gas = $59.0^oC=273+59.0=332K$

Now put all the given values in the above equation, we get:

$\frac{0.892atm\times 6.35L}{328K}=\frac{1.05atm\times V_2}{332K}$

$V_2=5.46L$

Thus, the final volume of gas will be, 5.46 L

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3 years ago

Pls help 25 pts

mote1985 [20]

Answer:

In the quantum-mechanical model of an atom, electrons in the same atom that have the same principal quantum number (n) or principal energy level are said to occupy an electron shell of the atom. Orbitals define regions in space where you are likely to find electrons.

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2 years ago