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3 years ago

How many atoms are there in 144g of sucrose, C12H22O11 or sugar molecules?

Chemistry

1 answer:

kati45 [8]3 years ago

8 0

Answer:

2.53×10²³ atoms

Explanation:

To solve this, convert grams to moles and moles to atoms.

To convert grams to moles, use the molar mass of C₁₂H₂₂O₁₁ (342.296 g/mol). Divide the mass by the molar mass to get moles. Moles = 0.4207

To convert moles to atoms, use Avogadro's number (6.022×10²³ atoms/mol). Multiply moles by Avogadro's number to get atoms. Atoms = 2.53×10²³

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Use significant digits to express the weight of a 150.0 lb person in kilograms ( 1 lb = 453.6 g)

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Answer:

150.0 = 68.038856

Explanation:

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2 years ago

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3 years ago

Can pressure affect the solubility of a substance in solution ? Explain

Tema [17]

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3 years ago

What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S

Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

<u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

c= Heat Capacity of Liquid= 4.184 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 $\frac{J}{gC}$ × 185.5 g× (- 25.6 °C)

Solving:

<u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× $\frac{1mol}{18 grams}$ = 10.30 moles, where 18 $\frac{g}{mol}$ is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 $\frac{kJ}{mol}$

Replacing:

Q2= 10.30 moles×6.01 $\frac{kJ}{mol}$

Solving:

<u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

c = Heat Capacity of Solid = 2.092 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 $\frac{J}{gC}$ × 185.5 g× (-10.70) °C

Solving:

<h3>Total heat required</h3>

The total heat required is calculated as:

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

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7 0

2 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0

1 year ago