All categories

3 years ago

Why potasium more reactive than lithium and sodium

1 answer:

5 0

So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom. Hope this answers the question. Have a nice day. Feel free to ask more questions.

You might be interested in

Calculate the moles of calcium chloride (CaCl2) needed to react in order to produce 85.00 grams of calcium carbonate (CaCO3). us

BartSMP [9]

Answer:

0.85 mole

Explanation:

Step 1:

The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:

When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:

CaCl2 + Na2CO3 -> CaCO3 + 2NaCl

Step 2:

Conversion of 85g of CaCO3 to mole. This is illustrated below:

Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

Mass of CaCO3 = 85g

Moles of CaCO3 =?

Number of mole = Mass /Molar Mass

Mole of CaCO3 = 85/100

Mole of caco= 0.85 mole

Step 3:

Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.

This is illustrated below :

From the balanced equation above,

1 mole of CaCl2 reacted to produced 1 mole of CaCO3.

Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.

From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3

5 0

3 years ago

Read 2 more answers

Sea water is an electrolyte.<br> True<br> False

antoniya [11.8K]

If you are talking about just pure regular water, the answer is false. BUT, some salts dissolved IN WATER, can act as electrolytes. But regular water, no.

8 0

3 years ago

Read 2 more answers

Which is less likely to be a relatable source of information the webpage of a university or the webpage a scientist who is tryin

faltersainse [42]

Answer:

The web page of a university

Explanation:

A scientist can be more biased within coming to information about pretty much anything. I have had multiple science teachers who seem more biased on to something else and pretend that they're right just cause they know what they are doing.

Then the university would be a great choice because its controlled by a higher state, then also the consistency of being updated.

7 0

3 years ago

Plz help me with 18 guys tysm!

kap26 [50]

A convergent boundary is when two plates push against each other. a transform boundary is when they slide past each other. both can couse mid ocean ridges.

5 0

3 years ago

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .