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2 years ago

List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)

Chemistry

1 answer:

ValentinkaMS [17]2 years ago

8 0

Answer: There are 20 carbons atoms, 62 oxygen atoms and 44 carbon atoms on the left side of the equation.

Explanation:

In the given chemical reaction:

$2C_{10}H_{22}(l)+31O_2(g)\rightarrow 20CO_2(g)+22H_2O(g)$

Reactants side = Left side

Number of atoms =

Coefficient × Number of atoms of an element in a unit molecular formula

Number of carbon atoms = 2 × 10 = 20

Number of oxygen atoms = 31 × 2 = 62

Number of hydrogen atoms = 2 × 22 = 44

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Which number is equal to –906,060?

mamaluj [8]

In absolute value it is equal to 906,060
In scientific notation it is equal to -9.06,060^5

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1 year ago

Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

Activity Two: Just Lemons, Inc. Production

Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.

Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients.

Water 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

Just Lemons Lemonade Recipe Equation:

2 water + sugar + lemon juice = 4 lemonade

Mole conversion factors:

1 mole of water = 1 cup = 236.59 g

1 mole of sugar = 1 cup = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

The Sun is a main sequence star. During the last stage of its life cycle, it will become a white dwarf. It will shrink in size,

dangina [55]

The one that is tiny and white.

3 0

1 year ago

Read 2 more answers

A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f

swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0

2 years ago

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

2 years ago