Answer:
97 J
Explanation:
Step 1: Given data
- Mass of the sample (m): 12 kg
- Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)
- Initial temperature: 45 K
Step 2: Calculate the temperature change
ΔT = 80 K - 45 K = 35 K
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J
A pure substance has a constant composition and cannot be separated into simpler substances by physical means. There are two types of pure substances: elements and compounds. Elements: are pure substances made up of only l type of atom. Atoms of the same element are identical in properties.
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
Answer:
E) 1, 2, and 3
Explanation:
50g H2O + 0.45g NaCl --> 50.45g saline solution
