All categories

2 years ago

How does a lithium cation compare to a lithium atom?

2 answers:

6 0

The correct answer is gonna be C) A lithium cation ion is smaller

A lithium cation has lost its valence electrons, which causes the remaining electrons to be pulled in stronger by the positive charge in the nucleus. As they get closer to the nucleus, the overall size of the atom is decreased.

Assoli18 [71]2 years ago

3 0

The correct answer is c

You might be interested in

A 8L sample of gas at 250 K is cooled to 75 K. What is the volume of the gas after it is cooled?

Stells [14]

$by \: Charles'\: law: \: \\ \frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}} \\$

Where v is the volume(in L) and t is the temperature(in °K)

$\frac{8}{250} = \frac{v_{2}}{75} \\\\4/5 =v_2/3 \\\\\boxed{v_{2} = \frac{12}{5}}\\\\\\\huge{=2.5 l}$

7 0

1 year ago

Where is most of the mass found in an atom

IgorLugansk [536]

massive livand that sarah or someone is how u do it

7 0

2 years ago

Read 2 more answers

Structural formula for:<br>5 - methyl hept - 3 - enal

leonid [27]

Answer:

draw the carbon chain is containing 6 carbon then attach the aldehyde group with edge carbon in chain then put the put double bond at 3 no. carbon

6 0

2 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

2 years ago

What did Kwang Jeon discover when researching amoeba infected by bacteria?

defon

<span>Kwang Jeon observed that Amoeba had been attacked by a bacterial infection, and lots of the Amoeba had died. However, some survived and continued to reproduce. After investigating the remaining Amoeba and their offspring, he noticed they were very healthy. He thought maybe they were able to fight off the bacteria, but instead, he found they were still infected with the bacteria but were not dying. The bacteria were no longer making the Amoeba sick. Then, he killed off the bacteria using antibiotics and was surprised to see that the Amoeba also died. It seemed the Amoeba and bacteria had formed a relationship in which they both needed each other to survive. After researching, Jeon found that the bacteria made a protein that the Amoeba needed to survive. </span>

6 0

2 years ago