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4 years ago

How many moles of h2so4 are needed to make 85 ml of .750 m solution?

Chemistry

1 answer:

harkovskaia [24]4 years ago

5 0

Answer:

0.06375 moles

Explanation:

Molarity is defined as moles of a solute per volume of solution.

-Given the molarity is 0.750m and the volume is 85ml

#Apply the molarity formula to solve for moles:

$molarity=\frac{moles}{volume, L}\\\\\therefore Moles=Volume, L\times Molarity\\\\\\=\frac{85}{1000}L\times 0.750\ mol/L\\\\=0.06375\ mol$

Hence, 0.06375 moles of sulfuric acid are needed.

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A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) doe

Sholpan [36]

P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g

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4 years ago

Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi

nevsk [136]

Explanation:

Haemoglobin consists of heme unit which is comprised of an $Fe^{2+}$ and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - Heme

The organic ring component of heme is - Porphyrin

Under normal conditions, the central atom of heme is - $Fe^{2+}$

In deoxyhemoglobin , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has six bonds: four to nitrogen atoms in the porphyrin, one to a histidine residue, and one to oxygen.

7 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

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3 years ago

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