Can I get an answer check on question one? Experiment 14Advance Study Assignment: Heat Effects and Calorimetry1. A metal sample weighing 147.90 g and at a temperature of 99.5°C was placed in 49.73 g of water in a calo-rimeter at 23.0°C. At equilibrium the temperature the water and metal was 41.8°C.What was Ar for the water? (At=final initial)41.8°C - 250°Ca.b. What was Ar for the metal?418-995d. Calculate the specific heat of the metal, using Equation 3.94-₂0=8H₂₂₂ Ato = - SH m Mm Atm9-4.18)(147.90)(-577)e. What is the approximate molar mass of the metal? (Use Eq. 4.)25MM-SH5C) (55,6714) = 0007008नाम)How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g °C.)9=SH; MART ⇒ (4,18³) (49.73.8) (18.88)Sb. Calculate 20, using Equation 1.Why?18,8c. Find AH for the reaction as it occurred in the calorimeter (Eq. 5).AH-577 d°C°C3908-4g/mol2. When 4.98 g of NaOH was dissolved in 49.72 g of water in a calorimeter at 23.7°C, the temperature of thesolution went up to 50.1°C.a. Is this dissolution reaction exothermic?joules35,671.4 joules/g cC7.01 X10joulesioulas

As you have asked for question one we are providing you the solution for the same. “Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts for you. To get the remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.” According to the question we have, The mass of the metal sample is given by = 147.90 g The mass of the water in the calorimeter is given by = 49.73 g The initial temperature of the metal sample is given by = 99.5°C The initial temperature of the water in the calorimeter is given by = 23.0°C The final temperature in the calorimeter at equilibrium is given by = 41.8°Ca) The change in the temperature ∆T of the water can be calculated by the given formulae, ∆T = Tfinal - Tinitial As we have both temperatures, we get ∆T = 41.8°C - 23.0°C=18.8°C The required change in temperature ∆T for the water is found to be 18.8°Cb) The change in the temperature ∆T for the metal can be calculated by, ∆T = Tfinal - Tinitial As we have both temperatures, we get ∆T = 41.8°C - 99.5°C=-57.7°C The required change in temperature ∆T for the metal sample is found to be -57.7°Cc) The heat (q) flow into the water can be calculated by, q=mC∆T where m is the mass of the water (49.73) and ∆T is the change in temperature for the water (18.8°C) C is the specific heat capacity for the water which is given by = 4.18 J/g°C Substituting all the values we get, q=49.73 g ×4.18 Jg °C×18.8°C=3907.98 J The required heat (q) flow into the water is found to be 3907.98 joules.