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2 years ago

I am giving brainly to the fastest

Chemistry

2 answers:

Svetllana [295]2 years ago

6 0

North Dakota and Michigan because it has to be 32 Farenhieht or less to snow

Feliz [49]2 years ago

4 0

Answer:

its the third one North Dakota and Michigan because it has to be 32 Farenhieht or less to snow Explanation:

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Dumbledore decides to gives a surprise demonstration. He starts with a hydrate of Na2CO3 which has a mass of 4.31 g before heati

vovikov84 [41]

Answer:

Na₂CO₃.2H₂O

Explanation:

For the hydrated compound, let us denote is by Na₂CO₃.xH₂O

The unknown is the value of x which is the amount of water of crystallisation.

Given values:

Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g

Mass after heating (Na₂CO₃) = 3.22g

Mass of the water of crystallisation = (4.31-3.22)g = 1.09g

To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:

Number of moles = $\frac{mass }{molar mass }$

Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹

Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹

Number of moles of Na₂CO₃ = $\frac{3.22}{106}$ = 0.03mole

Number of moles of H₂O = $\frac{1.09}{18}$ = 0.06mole

From the obtained number of moles:

Na₂CO₃ H₂O

0.03 0.06

Simplest

Ratio 0.03/0.03 0.03/0.06

1 2

Therefore, x = 2

7 0

3 years ago

Which of the following has higher ionization energy - Be or Ca. Explain your choice in terms of attractive force between protons

Vlad [161]

Answer:

Ca has the greater Ionization Energy because the Trend is I/e decreases as you move down a group. Therefore, Ca has the greater I/e

Explanation:

3 0

3 years ago

What is the electronic structure of carbon

horsena [70]

Answer:

[He] 2s2 2p2

Explanation:

5 0

3 years ago

Identify which one is the reducing agent in this reaction

Romashka-Z-Leto [24]

Answer:

Na.

Explanation:

The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced.

A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction.

For the reaction:

Na is oxidized to Na⁺ in (Na₂S) (loses 1 electron). "reducing agent".

S is reduced to S²⁻ in (Na₂S) (gains 2 electrons). "oxidizing agent".

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

2 years ago