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3 years ago

If one body is positively charged and another body is negatively charged, free electrons tend to

Chemistry

2 answers:

bulgar [2K]3 years ago

6 0

Answer:

A. move from the negatively charged body to the positively charged body.

Explanation:

because this is the correct answer

Naily [24]3 years ago

4 0

I think the correct answer is A. If one body is positively charged and another body is negatively charged, free electrons tend to <span>move from the negatively charged body to the positively charged body. This happens because they have different charges and therefore electrons are more attracted to the positive or its opposite charge.</span>

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HELP!!! I need "2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)" in word form

baherus [9]

I don’t get it what do u need exactly?

8 0

3 years ago

1. write the complete chemical symbol for the ion with 14 protons 15 neutrons and 18 electrons

motikmotik

Answer:

1. $Si^{-4}$

2. $Co^{+2}$

Explanation:

An ion is formed when an atom that is said to be neutral gains or losses electrons.

It is thought that a negative ion (anion) is produced as it gains electrons and a positive ion (cation) is formed when it loses an electron.

Atomic number is the total number of protons and electrons in a neutral atom.

From the information

Protons = 14

electron = 18

Net Charge = no of proton - no of electron

= 14 - 18 = -4

Mass number = 14 + 15 = 29

Thus, the chemical symbol = $Si^{-4}$

For ion with 27 proton, 32 neutrons and 25 electrons

Net charge = 27 - 25 = +2

Mass number = 27 + 32 = 59

Thus, the chemical symbol = $Co^{+2}$

6 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago

If 4.5 mol of C6H8O7 react, how many moles of CO2 and Na3C6H5O7 will be produced?

Salsk061 [2.6K]

Answer:

A. 13.5 moles of CO2

B. 4.5 moles of Na3C6H5O7

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3NaHCO3(aq) + C6H8O7(aq) → 3CO2(g) + 3H2O(s) + Na3C6H5O7(aq)

A. Determination of the number of mole CO2 produced from the reaction. This is illustrated below:

From the balanced equation above,

1 mole of C6H8O7 reacted to produce 3 moles of CO2.

Therefore, 4.5 moles of C6H8O7 will react to produce = 4.5 x 3 = 13.5 moles of CO2.

From the above calculation, 13.5 moles of CO2 is produced

B. Determination of the number of mole of Na3C6H5O7 produced by the reaction. This is illustrated below:

From the balanced equation above,

1 mole of C6H8O7 produced 1 mole of Na3C6H5O7.

Therefore, 4.5 moles of C6H8O7 will also produce 4.5 moles of Na3C6H5O7.

Therefore, 4.5 moles of Na3C6H5O7 is produced from the reaction.

5 0

3 years ago

Which best describes the effect of J. J. Thomson’s discovery?

worty [1.4K]

Answer:

A. The accepted model of the atom was changed.

Explanation:

Thomson described an atom that has the protons uniformly distributed within a circular atom with the electrons embedded . However, it cannot explain the emission of electrons from the surfaces of metals like photoelectric emission. it does not explain how the particle of the atom contribute significantly to the entire mass of an atom

7 0

4 years ago