Answer:
- The density of ethanol measures 0.80 g/ml.
Explanation:
The density of water is known to be 1 g/ml, indicating that 158 g of water occupies a volume of 158 ml.
d = m/v = 258/258 = 1 g/ml
This implies the volume will remain the same for ethanol.
hence, the density of ethanol = 127/158 = 0.80 g/ml
<span>Reaksi antara besi dan asam klorida menghasilkan besi (II) klorida serta gas hidrogen.</span>
Each isotope contains an identical number of protons and electrons, but they vary in neutron count, making them isotopes.
To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.
Answer:
To achieve the desired outcome, 8.55 mL of NaOH is necessary.
Explanation:
Considering that:
the weak acid has a mass of 0.4 g
and a molecular weight of 234 g/mol
so, the number of moles of the weak acid is calculated as 0.4 g/234 g/mol = 0.00171 mole
To convert half of the weak acid (WA) to conjugate base (CB), we must add NaOH.
Thus, [WA]=[CB] 0.00171/2 = 8.55×10⁻⁴ mole of NaOH required
Further, knowing that the concentration of NaOH is 0.10 M
the volume needed to achieve this result can be calculated as follows:
= 
= 8.55 mL of NaOH is necessary
