The answer is 308 K.
The formula is C + 273.15 = K
this meaning the formula for this problem would be 35 + 273.15 = 308.15.
Out of all the option choices that would be rounded to 308K.
6. a) Team 3 (b/c their numbers were closer apart then the rest)
b)No they were not (can happen you might be precises but not accurate)
c)Team four, (on trial 4)
Answer:
NaNO₃ and AgCl are the two products that can be formed.
Sodium nitrate, an aqueous solution and a solid silver chloride (precipitate)
Explanation:
We determine the dissociation of both salts
AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq)
NaCl (aq) → Na⁺ (aq) + Cl⁻ (aq)
We make the ionic equation:
Ag⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) → NaNO₃(aq) + AgCl (s) ↓
Answer:
Absolute zero temperature: Absolute zero is the temperature at which a substance have very low internal energy or in other words no heat energy in the particle.
Explanation: The temperature is very low and the particles are very cold. On cooling, speed of particle decrease. On the Celsius scale, - 273.15 is the absolute zero and on the Kelvin scale, 0 k is the absolute zero. On absolute zero temperature, the movement in the particles occurs at very low speed near to zero.
Answer:
(a) 282 kJ
(b) 67.4 Calories
Explanation:
(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.
We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):
(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ
(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.
(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories
