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3 years ago

In a classroom which comparison would a teacher most likely use for describing a mole

1 answer:

3 0

Mole - one of the most important concepts in chemistry - is a kind of link to go from the microworld of atoms and molecules in a normal macrocosm grams and kilograms.
In chemistry often have to consider large numbers of atoms and molecules. For fast and efficient calculation made using the weighing method. But it is necessary to know the weight of individual atoms and molecules. In order to identify the molecular weight must be added the weight of all atoms in the compound.

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Consider this reaction:

ryzh [129]

Answer:

I think it's 6 moles are produced

3 0

2 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago

Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c

garri49 [273]

The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.

5 0

3 years ago

Read 2 more answers

Britney added 0.05 moles of copper(II) nitrate solution to 0.1 moles of sodium hydroxide solution and

Rama09 [41]

The percent yield of copper hydroxide is 84%

<h3>Stoichiometry</h3>

From the question, we are to determine the percent yield of copper hydroxide

First, we will determine the theoretical mass

From the given balanced chemical equation, we have

Cu(NO₃)₂ + 2NaOH -- Cu(OH)₂ + 2NaNO₃

This means,

1 mole of copper(II) nitrate reacts with 2 moles of sodium hydroxide to produce 1 mole of copper hydroxide

Therefore,
0.05 mole of copper(II) nitrate reacts with 0.1 mole of sodium hydroxide to produce 0.05 mole of copper hydroxide

The theoretical number of moles of copper hydroxide that is produced is 0.05 mole

Now, for the theoretical mass

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of copper hydroxide = 97.56 g/mol

Then,

Theoretical mass = 0.05 × 97.56

Theoretical mass of copper of hydroxide produced is = 4.878 g

Now, for the percent yield of copper hydroxide

Percent yield is given by the formula,

$Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

Then,

$Percent\ yield\ of\ copper\ hydroxide= \frac{4.1}{4.878}\times 100\%$

$Percent\ yield\ of\ copper\ hydroxide= 84\%$

Hence, the percent yield of copper hydroxide is 84%.

Learn more on Stoichiometry here: brainly.com/question/9372758

7 0

2 years ago

8. If a chemical reaction such as photosynthesis begins with 6 atoms of carbon (C), how many atoms of carbon (C) should be in th

NISA [10]

Answer

Explanation:

4 0

2 years ago