Consider the following equation CO plus 2 H2 and acirc and 134 and 146 CH3OH and acirc and 150 and sup3H rxn -128 kJ Calculate the amount of heat (in kJ) associated with complete reaction of 8.08 g H2?
To find the freezing point of the solution, we first calculate the molality (m) of the LiBr solution. Since 0.5 mol of LiBr is dissolved in 0.5 kg of water (500 mL of water), the molality is 1.0 m. Using the formula for freezing point depression, ΔTf = Kf * m, where Kf = 1.86 °C/m, we get ΔTf = 1.86 °C/m * 1.0 m = 1.86 °C. Thus, the freezing point of the solution is 0 °C - 1.86 °C = -1.86 °C.
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To calculate the heat transferred (Q) to the aluminum pizza pan, we can use the formula: [ Q = m \cdot c \cdot \Delta T ] where ( m = 480 , \text{g} ), ( c = 0.96 , \text{J/g°C} ), and ( \Delta T = 234°C - 22°C = 212°C ). Substituting the values: [ Q = 480 , \text{g} \cdot 0.96 , \text{J/g°C} \cdot 212°C \approx 97,036.8 , \text{J} ] To convert to kilojoules, divide by 1000: [ Q \approx 97.04 , \text{kJ} ] Thus, approximately 97.04 kJ of heat must be transferred.
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The word in bold that means "capable of being easily influenced" is pliant. This word retains the root meaning of "to bend" from the Old French root plier, suggesting a sense of flexibility or adaptability in being influenced.
118 degrees
You pay $21 for 3 ½ tons of coal. What will be the bill for 7 ½ tons?
Without any equality sign and the given terms in disarray it can't be considered to be an equation and so therefore it follows that no solution is possible.
core values
Consider the following equation CO plus 2 H2 and acirc and 134 and 146 CH3OH and acirc and 150 and sup3H rxn -128 kJ Calculate the amount of heat (in kJ) associated with complete reaction of 8.08 g H2?
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The weather report inaccurately stated the temperature range as between 40 and 50 degrees instead of the actual 54.4 degrees. This means the weather report incorrectly suggested the temperature was lower than it actually was.
To find the freezing point of the solution, we first calculate the molality (m) of the LiBr solution. Since 0.5 mol of LiBr is dissolved in 0.5 kg of water (500 mL of water), the molality is 1.0 m. Using the formula for freezing point depression, ΔTf = Kf * m, where Kf = 1.86 °C/m, we get ΔTf = 1.86 °C/m * 1.0 m = 1.86 °C. Thus, the freezing point of the solution is 0 °C - 1.86 °C = -1.86 °C.