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3 years ago

Classify chunky peanut butter as an element, a compound, a heterogeneous mixture, or a homogeneous mixture.

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Chunky peanut butter is a mixture which is NOT uniform throughout, so it is a heterogeneous mixture.

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When elements form bonds, it changes their ______ properties. The answer is..?

Arturiano [62]

Answer:

chemical and physical

6 0

3 years ago

Calculate ΔHrxnΔHrxn for the following reaction: 3C(s)+4H2(g)→C3H8(l)3C(s)+4H2(g)→C3H8(l) Use the following reactions and given

mihalych1998 [28]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -120.9 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$3C(s)+4H_2(g)\rightarrow C_3H_8(l)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C_3H_8(l)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$ $\Delta H_1=-2026.6kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$ ( × 3)

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$ $\Delta H_2=-483.5kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[3\times \Delta H_2]+[2\times \Delta H_3]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -(-2026.6))+(3\times (-393.5))+(2\times (-483.5))]=-120.9kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -120.9 kJ.

8 0

3 years ago

Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and

notka56 [123]

Answer:

$pH=4.56$

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

$n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol$

And the concentrations are:

$[acid]=0.000855mol/0.025L=0.0342M$

$[base]=0.000781mol/0.025L=0.0312M$

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

$pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56$

Best regards!

6 0

2 years ago

What electron could have quantum numbers n = 4, l = 2, ml = -2, ms = -?

brilliants [131]

The answer is ms= -1/2

8 0

3 years ago

A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?

Oksana_A [137]

Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

$\frac{V1}{T1} = \frac{V2}{T2} \\ \\ T2= \frac{V2*T1}{V1}= \frac{235 mL * 105 ^{\circ}C }{455 mL}=54,23 ^{\circ}C$

So, the final temperature is 54,23 °C

Have a nice day!

5 0

3 years ago

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