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3 years ago

State one source of lead compounds in the air?

1 answer:

7 0

Answer:Sources of lead emissions vary from one area to another. At the national level, major sources of lead in the air are ore and metals processing and piston-engine aircraft operating on leaded aviation fuel. Other sources are waste incinerators, utilities, and lead-acid battery manufacturers. The highest air concentrations of lead are usually found near lead smelters.

Explanation:

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What two things must you check when seeing if a lewis structure is correct

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Answer:

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3 years ago

oxidul de cupru este folosit la colorarea sticlei si a emailurilor carora le imprima o culoare verde-albastruie.Calculeaza masa

Pani-rosa [81]

Answer:

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3 years ago

0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

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3 years ago

__________ is the use of high energy radiation to kill cancer cells.

Ainat [17]

Answer:

Radiation therapy is the use of high energy radiation to kill cancer cells.

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2 years ago

Read 2 more answers

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the max

jolli1 [7]

Answer: The mass of aluminium chloride that can be formed are 46.3 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Aluminium:

Given mass of aluminium = 32 g

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol$

For Chlorine:

Given mass of chlorine = 37 g

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:

$\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol$

For the given chemical equation:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

$0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g$

Hence, the mass of aluminium chloride that can be formed are 46.3 g

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4 years ago