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3 years ago

13

Which family in the periodic table has metals that are usually harder than other metals? Alkali metals Alkaline earth metals Lan

thanide series Transition metals

1 answer:

stealth61 [152]3 years ago

3 0

Answer:

The correct answer is "Alkaline earth metals".

Explanation:

Alkaline earth metals are a group of elements that are located in group 2 of the periodic table. They are harder than alkaline metals, have brightness and are good electrical conductors. Hardness is the degree of scratch resistance a material offers.

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0.00000000082 -<br> scientific notation

Nadusha1986 [10]

Answer:

8.2 x 106^-11

Explanation:

To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2

Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.

Once the decimal is in the right spot count the spots moved.

Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.

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3 years ago

What is the purpose of finding oxidation states in the half-reaction method for balancing equations?

vazorg [7]

Answer:

B. To Identify the half reactions for the equation

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3 years ago

Is chromium a transition metal

oksano4ka [1.4K]

No it is not

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3 years ago

Read 2 more answers

Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl

stich3 [128]

Answer: Molarity of chloride anion = 0.32 M

<em>Note: the question is missing some values. The full question is given below;</em>

<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>

Explanation:

Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M

The mass of nickel (II) chloride is 7.26 g.

The volume of potassium carbonate is 350 mL = 0.35 L

The molarity of potassium carbonate solution is 0.50 M

The reaction of nickel (II) chloride and potassium carbonate is given below.

NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)

The dissociation of nickel (II) chloride is given below.

NiCl₂ -----> Ni²⁺ + 2Cl⁻

The molar mass of nickel (II) chloride is 129.6 g/mol

The moles of nickel (II) chloride can be calculated by the formula given below;

No of moles = mass(g) / molar mass (g/mol)

No of moles = 7.26 / 129.6 = 0.056 moles

Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M

The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.

Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M

3 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