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3 years ago

How many milliliters of .085 m naoh are required to titrate 25 ml of .072 m hbr to the equivalence point?

1 answer:

8 0

The ML of 0.85 m NaOH required to titrate 25 ml of 0.72m hbr to the equivalence point is calculated as follows

calculate the moles of HBr used

moles = molarity x volume

25 x0.072/1000= 0.0018 moles

write the equation for reaction

NaOH + HBr = NaBr + H2O
from reacting equation the mole ratio between NaOH to HBr is 1:1 therefore the moles of NaOH = 0.0018 moles

volume = moles/molarity
0.0018/0.085 = 0.021 L in Ml = 0.021 x1000=21.18 Ml ofNaOH

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You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

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

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

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

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$

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3 years ago

Can anyone help me with a chemistry problem please?

steposvetlana [31]

Answer:

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2 years ago

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When one evaporates a pure substance, what would be seen after this evaporation?

satela [25.4K]

Answer:

I would like to understand the temperature at which a substance will vaporize when dissolved in a liquid. I have researched this online for hours, but haven't found a conclusive answer. Is it the boiling point of the dissolved substance? I'm attempting to find the temperature at which caffeine vaporizes when dissolved in water or other vegetable glycerin.

Explanation:

I hope this helps a little bit

4 0

3 years ago

A 500.0 g block of dry ice (solid CO2, molar mass = 44.0 g) vaporizes at room temperature. Calculate the volume of gas produced

Damm [24]

Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Ideal gas law</h3>

An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Volume of gas</h3>

In this case, you know:

P= 1.50 atm
V= ?
n= 500 g× $\frac{1 mole}{44 g}$ = 11.36 moles, being 44 $\frac{g}{mole}$ the molar mass of CO₂
R= 0.082 $\frac{atmL}{molK}$
T= 25 C= 298 K (being 0 C=273 K)