Answer:
38.9 grams of 
Explanation:
0.187 mol BaCl2 x 
0.187 m x 208 g/m
0.187 x 208 g
38.896 g --> 38.9 g BaCl2
Given :
Volume , V = 500 mL .
Molarity , M = 0.5 M .
Molecular mass of NaCl is 
.
To Find :
How many grams of NaCl is required .
Solution :
Let , NaCl required is x gram .
Molarity is given by :
Hence , this is the required solution.
During cellular respiration, the carbon and hydrogen atoms change partners and bond with oxygen atoms instead. The carbon-hydrogen bonds are replaced by carbon-oxygen and hydrogen-oxygen bonds. As the electrons of these bonds "fall" toward oxygen, energy is released.
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
