Answer:
0.007 M
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
The expression of the pH of the calculation of weak acid is:-
Where, C is the concentration = ?
Given, pH = 3.45
So, for 
, 
C = 0.007 M
Answer:
35.8 g
Explanation:
Step 1: Given data
Mass of water: 63.5 g
Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C
Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.
63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl
So let's use some equations to represent the data [let R= cost of ring & B= cost of bracelet]
R= B + $ 36 .... (1)
B=
× R ... (2)
By using simultaneous equations to solve for B and R.
Substitute eq. (1) into eq. (2)
B =
× (B + $36)
B =
B +
⇒ B = $48
By substituting value of B into ea (1)
If R = B + $36
R = ($48) + $36
= $84
∴ <span> the total of the two items = R + B
= $84 + $48
</span> = $132
Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
1. Both part of the ecosystem
2. There are biotic objects on abiotic objects ( caterpillars on trees ) and abiotic objects on biotic things ( pollen on bees )
3. Both are made of atoms
