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Romashka-Z-Leto [24]

3 years ago

6

In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After

dissolution of the salt, the final temperature of the calorimeter contents is 23.34 degrees C. Assuming the solution has a heat capacity of 4.18 J/Cg and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

1 answer:

igomit [66]3 years ago

7 0

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

<em>+ because the heat is absorbed, the reaction is endothermic-</em>

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Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas

Wittaler [7]

To determine the absolute pressure of this gas, all you need to do is to add the value of atmospheric pressure and the value of gage pressure.

Atmospheric pressure is equivalent to 100 kPa.
Gage pressure is 276 kPa.

Then, we add both values.

N = 100 kPa + 276 kPa
N = 376 kPa

The absolute pressure of this gas is 376 kPa.

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3 0

4 years ago

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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B

loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is present in $H_2O$ , Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

= 0.2043 moles

∴ moles of C = 0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

= 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

= 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H = 0.327 moles

molar mass of H = 1.008 g/mol

∴ mass of H in the sample = 0.327 moles × 1.008 g/mol

= 0.329616g

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C H O

0.2043 0.327 0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$ $\frac{0.327}{0.0136}$ $\frac{0.0136}{0.0136}$

15.02 24.04 1

≅

15 24 1

Therefore, the empirical formula would be : $C_{15}H_{24}0$

7 0

3 years ago

a light ray hits a hard, opaque object. the angle at which the light ray bounces off the object depends on the smoothness of the

kumpel [21]

The answer would be the third option. (the angle at which the light hits the surface.)<span>
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3 years ago

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How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh

zysi [14]

Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

CO + 2 H₂   → CH₃OH

Calculating Moles of CO:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO

Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g

X = 11.25 Moles of CO

Calculating Moles of H₂:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂

Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g

X =  22.5 Moles of H₂

Result:
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