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2 years ago

What are the correct units for molar mass? a. b. mol C. g/mol d. g*mol

Chemistry

2 answers:

yawa3891 [41]2 years ago

8 0

Answer:

the correct unit for molar mass is g/mol

Drupady [299]2 years ago

3 0

Answer:

The molar mass of a compound tells you the mass of 1 mole of that one substance. Tt tells you the number of grams per mole of a compound. The units for molar mass are, therefore, grams/mole.

Explanation:

BOOM BIG BRAIN

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PLEASE ANSWER AS QUICKLY AS POSSIBLE THANK YOU SO MUCH

Harlamova29_29 [7]

Answer:

This description needs a unit describing the system of measurement.

6 0

1 year ago

Sodium is a silvery-white metal that reacts with chlorine gas, which is a yellow-greenish gas that is toxic. The reaction gives

Amiraneli [1.4K]

Answer:

Sodium is a silvery-white metal that reacts with chlorine gas, which is a yellow-greenish gas that is toxic. The reaction gives off a lot of heat. After the reaction, which statement is true about the chemical properties of the product, sodium chloride?

Pure sodium reacts violently and sometimes explosively with water producing sodium hydroxide, hydrogen gas and heat

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Chlorine is a very poisonous yellow green gas with a sharp odour that was used in gas warfare during WW1

Sodium and chlorine reacts with each other, however, to produce one of the most familiar substance used in cooking and preservation industry today Sodium Chloride or Common salt or table salt in the irreversible equation;

2Na(s) + Cl2(g) → 2NaCl(s)

Explanation:

It is easy to see why this reaction takes place so readily sodium has one electron in its outermost valence shell while chlorine has seven electrons in its valence shell. when sodium atom transfers one electron to chlorine atom forming a sodium cation (Na+) and a chloride anion (Cl-) both ions have complete valence shells and are energetically more stable. the reaction is extremely exothermic, producing a bright yellow light and a great deal of heat and fumes of sodium chloride.

In a reaction observation of the reaction process you will see sodium flares up almost immediately upon reaction with water.

3 0

3 years ago

Which half-reaction correctly shows the oxidation of iron

pishuonlain [190]

Answer:

What is the oxidation half reaction for iron?

The two elements involved, iron and chlorine, each change oxidation state; iron from +2 to +3, chlorine from 0 to -1. There are then effectively two half-reactions occurring. These changes can be represented in formulas by inserting appropriate electrons into each half-reaction: Fe2+ → Fe3+ + e.

Hope this helps..

8 0

2 years ago

Which characteristic of minerals is represented in this picture?

Marianna [84]

B is the correct answer to your question

7 0

3 years ago

Read 2 more answers

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