The mass is 5kg
I hope this helps
he value for the equilibrium constant for the following chemical reaction, the auto-ionization of water, is 1.0x10-14 at 298 K <u>1x10-14 0.5x10-14 2x10-14 -1x10-14 1x1014 1x10-15</u>
<h3>What is
chemical reaction?</h3>
A chemical reaction is a procedure that causes one group of chemical components to change chemically into another. Chemical reactions, which can frequently be described by a chemical equation, traditionally include changes that only affect the locations of electrons in the formation and dissolution of chemical bonds between atoms, with no change to the nuclei (no change to the elements present). The study of chemical processes involving unstable and radioactive elements, where both electronic and nuclear changes may take place, is known as nuclear chemistry.
Reactants or reagents are the substance(s) or substances that are initially utilized in a chemical reaction.
To learn more about chemical reaction, from the given link:
brainly.com/question/14197404
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Answer:
Scientists attribute the global warming trend observed since the mid-20th century to the human expansion of the "greenhouse effect" — warming that results when the atmosphere traps heat radiating from Earth toward space.
Certain gases in the atmosphere block heat from escaping. Long-lived gases that remain semi-permanently in the atmosphere and do not respond physically or chemically to changes in temperature are described as "forcing" climate change. Gases, such as water vapor, which respond physically or chemically to changes in temperature are seen as "feedbacks."
Explanation:
Answer:
![\Delta H_{f,C_3H_4}=276.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bf%2CC_3H_4%7D%3D276.8kJ%2Fmol)
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
![\Delta H_{rxn} =- m_wC_w\Delta T](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%20%3D-%20m_wC_w%5CDelta%20T)
We plug in the mass of water, temperature change and specific heat to obtain:
![\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%20%3D-%20%2835000g%29%284.184%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%20%29%282.316%5C%C2%B0C%29%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-339.16kJ)
Now, this enthalpy of reaction corresponds to the combustion of propyne:
![C_3H_4+4O_2\rightarrow 3CO_2+2H_2O](https://tex.z-dn.net/?f=C_3H_4%2B4O_2%5Crightarrow%203CO_2%2B2H_2O)
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
![\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D3%5CDelta%20H_%7Bf%2CCO_2%7D%2B2%5CDelta%20H_%7Bf%2CH_2O%7D-%5CDelta%20H_%7Bf%2CC_3H_4%7D)
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
![\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%20%3D-339.16kJ%2A%5Cfrac%7B1%7D%7B7.00g%7D%2A%5Cfrac%7B40.06g%7D%7B1mol%7D%3D-1940.9kJ%2Fmol)
Now, we solve for the enthalpy of formation of C3H4 as shown below:
![\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bf%2CC_3H_4%7D%3D3%5CDelta%20H_%7Bf%2CCO_2%7D%2B2%5CDelta%20H_%7Bf%2CH_2O%7D-%5CDelta%20H_%7Brxn%7D)
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
![\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bf%2CC_3H_4%7D%3D3%28-393.5kJ%2Fmol%29%2B2%28-241.8kJ%2Fmol%29-%28-1940.9kJ%2Fmol%29%5C%5C%5C%5C%5CDelta%20H_%7Bf%2CC_3H_4%7D%3D276.8kJ%2Fmol)
Best regards!