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3 years ago

Which one of the following is the net ionic equation for the reaction of nitric acid with aluminum hydroxide

1 answer:

5 0

The ionic eqn is as follow:

1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)

3moles of No3- ion on both sides cancels out to give the net ionic eqn:

1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)

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The leaves of plants contain specialized parts that absorb energy from sunlight. Which feature would help a plant

Ber [7]

Chlorophyll pigment found in chloroplast of leaves absorbs sunlight.

The plants has a technique in which they lean towards sunlight which helps them better to compete for sunlight.

Explanation:

The leaves of the plant consist of chloroplasts in which light absorbing pigment chlorophyll is found. The chloroplast consist of thylakoid where light reaction of photosynthesis takes place.

The presence of chlorophyll a in large amount would help plant better use the sunlight this amounts to increase number of chloroplast.

Leaning towards the sunlight is feature which helps them using more sunlight.

6 0

3 years ago

Read 2 more answers

Which of the following examples of erosion would occur most quickly?

anyanavicka [17]

Answer: OC

Explanation:

The Others Could Take Many Millions Of Years

8 0

3 years ago

I need the answer to 452 of arogon

yaroslaw [1]

I'm confused... is there more info?

3 0

3 years ago

What’s the reaction type of aqueous magnesium sulfate + aqueous sodium carbonate?

IrinaK [193]

Answer: Double displacement reaction

Explanation:

Double displacement reaction is defined as the reaction where exchange of ions takes place.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The chemical reaction between aqueous magnesium sulfate and aqueous sodium carbonate is represented as:

$MgSO_4(aq)+Na_2CO_3(aq)\rightarrow Na_2SO_4(aq)+MgCO_3(s)$

8 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago