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faltersainse [42]
3 years ago
10

In the boxes below, draw three particle diagrams representing the sample

Chemistry
1 answer:
Alexxandr [17]3 years ago
5 0

Answer: 2

Explanation: bencause they are changing

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25 points!!! please Which types of elements areconsidered neutraland Where are these types of elements found on the periodic tab
posledela

Elements which composed of an equal amount of three components protons, neutrons and electrons are considered neutral elements. example of neutral elements are hydrogen, helium, lithium and beryllium...

7 0
3 years ago
7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat
Pie

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

<h3>9.69g of S₂Cl₂</h3>
3 0
3 years ago
PLEASE HELP ASAP!
choli [55]

Answer:

\large \boxed{\text{D. 710 g}}

Explanation:

1. Calculate the molar mass of Na₂SO₄

\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}

The molar mass of Na₂SO₄ is 142 g/mol.

2. Calculate the moles of Na₂SO₄

\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}

3. Calculate the mass of Na₂SO₄

\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}

6 0
3 years ago
One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5
Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of Fe_2O_3 = 160 g/mole

First we have to calculate the moles of iron.

\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles

Now we have to calculate the moles of Fe_2O_3.

The balanced reaction is,

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of Fe_2O_3

So, 0.276 moles of iron obtained from \frac{0.276}{2}=0.138 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3

\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g

Therefore, the amount of Fe_2O_3 required are, 22.1 grams.

6 0
3 years ago
Which of the following options would be the best for dissolving PbBr2(s)?
Nezavi [6.7K]

Answer:

2) Add a solution of NaBr

Explanation:

Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr

PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr

Therefore, the best solution for dissolving PbBr₂(s) is NaBr

4 0
3 years ago
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