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3 years ago

In the space, show a correct numerical setup for calculating the number of moles of CO2 present in 11 grams of CO2

Chemistry

1 answer:

makvit [3.9K]3 years ago

4 0

Ans: Moles of CO2 = 0.25 moles

Given:

Mass of CO2 = 11 g

To determine

Moles of CO2

Explanation:

Moles = mass/molecular mass

Molecular mass of CO2 = 44 g/mol

Moles of CO2 = 11 g/44 g.mol-1 = 0.25 moles

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2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams of H2 are also produced?

faust18 [17]

So,

Our conceptual plan is as follows:

g AlCl3 --> mol AlCl3 --> mol H2 --> g H2

$\frac{129g\ AlCl_3}{133.4g/mol}\ *\frac{3mol\ H_2}{2mol\ AlCl_3}\ *2.016g/mol \\ \\=2.92 g\ H_2$

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3 years ago

Read 2 more answers

Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains

qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 650 mL

$\text{Mass of water}=1.00g/mL\times 650mL=650g$

Now we have to calculate the heat released on cooling.

Heat released on cooling = $m\times c\times (T_2-T_1)$

where,

m = mass of water = 650 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_2$ = final temperature = $29^oC$

$T_2$ = initial temperature = $0^oC$

Now put all the given values in the above expression, we get:

Heat released on cooling = $650g\times 4.18J/g^oC\times (29-0)^oC$

Heat released on cooling = 78793 J = 78.793 kJ (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = $\frac{\text{Total heat}}{\text{Heat for 1 ice cube}}$

Number of ice cubes melted = $\frac{78.793kJ}{6.02kJ}$

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

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3 years ago

What charge would you expect for an formed by S?

sp2606 [1]

Answer:

in this the correct answer is option 2.

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3 years ago

Injecting salt crystals over the ocean to grow cloud droplets has been proposed in efforts to make the clouds __________ thereby

anastassius [24]

Injecting salt crystals over the ocean to grow cloud droplets has been proposed in efforts to make the clouds brighter thereby affecting the radiation budget. The light of the sun shines on Earth, some of that light is reflected by the clouds back to space and some of the light reaches the earth and warms our planet. The earth and the hot oceans emit infrared radiation (IR), which we feel as heat. That IR "light"; returns to space through the atmosphere. Most are trapped by greenhouse gases, which keep the earth warm. Soon after, the IR radiation returns to space. Scientists call this "energy budget of the Earth" this cycle of incoming and outgoing energy.

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3 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

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

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

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 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

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3 years ago