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faltersainse [42]
3 years ago
8

Convert 15.78 g of H2O2 to mols. Convert 15.3 mols Au to g

Chemistry
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

.46 moles H2O2

3,014 grams Au

Explanation:

H2O2:

15.78g (1 mol/34g) = .46 moles H2O2

Au:

15.3mol (197g/mol) = 3,014 grams Au

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The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo
RSB [31]

Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹)  because it is  already a common factor to all.

The LCM  is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

4 0
3 years ago
Does 1 gram of phosphorus react with 6 grams of iodine to form 4 grams of phosphorus triodine in P4(s)+6I2(s)=4PI3(s)
mafiozo [28]

Answer:

No

Explanation:

One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.

When one gram phosphorus and 6 gram of  iodine react they gives 8.234 g ram of PI₃ .

Given data:

Mass of phosphorus = 1 g

Mass of iodine = 6 g

Mass of  PI₃ = ?

Solution:

Chemical equation:

P₄ + 6I₂    →  4PI₃

Number of moles of P₄:

Number of moles = Mass /molar mass

Number of mole = 1 g / 123.9 g/mol

Number of moles  = 0.01 mol

Number of moles of I₂:

Number of moles  = Mass /molar mass

Number of moles = 6 g / 253.8 g/mol

Number of moles = 0.024 mol

Now we will compare the moles of PI₃ with I₂ and P₄.

                I₂              :              PI₃

                  6              :               4

                 0.024       :             4/6×0.024 = 0.02

                  P₄            :               PI₃

                 1                :                4

                 0.01          :               4 × 0.01 = 0.04  mol

The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.

Mass of PI₃ = moles × molar mass

Mass of PI₃ = 0.02 mol × 411.7 g/mol

Mass of PI₃ =  8.234 g

4 0
3 years ago
The discipline of chemistry has made agricultural chemicals, such as fertilizer and pesticides, available to society.
vesna_86 [32]
Contributing to greenhouse gases
7 0
3 years ago
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Graphite, an allotrope of carbon, is converted into cubic diamond through a process that may take a billion years or longer.
andriy [413]

Answer:

The given statement is true.

Explanation:

The given statement is true.

As per an estimate, to convert graphite into diamond nearly 1 billion to 3.3 billion years of time is required.

Graphite converts into diamond when high-pressure is applied deep in the earth core.

5 0
3 years ago
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

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

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

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

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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