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3 years ago

Which formula correctly represents antinomy trichloride? SnS2 SnCl3 SbCl2 SbCl3

Chemistry

2 answers:

Lerok [7]3 years ago

4 0

Answer:

SbCl3

Explanation:

because Sb is antinomy and Cl3 stands for trichloride

max2010maxim [7]3 years ago

3 0

Answer:

SbcI3

Explanation:

The symbol of antimony is 'Sb'.

The symbol of chlorine is 'Cl'

First write down the symbol of the first element.

Use the prefix to determine the atoms of first element. If there is no prefix on element then there is only 1 atom.

Now write down the symbol of the second element.

Use the prefix to determine the atoms of second element.

Use prefix as 'mono' for '1', 'di' for '2', 'tri' for '3' and so on.

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Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones

Scilla [17]

Answer:

$_{97}^{345}\textrm{Bk}$

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

$^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is $_{97}^{345}\textrm{Bk}$

7 0

4 years ago

Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen

OLga [1]

Answer:

5.9 kg

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ: 55.84

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

m/kg: 7.0

(a) Moles of Fe

$\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}$

(b) Moles of CO

$\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}$

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

$\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}$

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ: 32.00

2C + O₂ ⟶ 2CO

n/mol: 261

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \text{4180 g O}_{2}$

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

$\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}$