Answer:
0.75 g/cm³
Explanation:
Given data:
Mass of wooden block = 180 g
Length of block = 10 cm
Width of block = 6 cm
Height or thickness = 4 cm
Density of block = ?
Solution:
Volume of block = height × length × width
Volume of block = 4 cm × 10 cm× 6 cm
Volume of block = 240 cm³
Density of block:
density = mass/ volume
d = 180 g/ 240 cm³
d = 0.75 g/cm³
Answer:
1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules
Explanation:
1. Mass / Molar mass = Mol
5g / 28 g/m = 0.178 moles
2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.
4x10²³ x2 = 8x10²³ atoms
3. 1 mol of anything, has 6.02x10²³ particles
6.02x10²³ molecules . 1.2 mol = 7.22x10²³
4. 1 atom of C weighs 12 amu.
4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu
1 amu = 1.66054x10⁻²⁴g
5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g
5. Molar mass NaCl = 58.45 g/m
1.3 g / 58.45 g/m = 0.0222 moles
1 mol has 6.02x10²³ atoms
0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²
6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water
Molar mass H₂O = 18 g/m
500 g / 18 g/m = 27.8 moles
6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵
Answer:
317.6 mL
Explanation:
Step 1: Write the balanced neutralization equation
MgO + 2 HCl ⇒ MgCl₂ + H₂O
Step 2: Calculate the mass corresponding to 640.0 mg of MgO
The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:
0.6400 g × (1 mol/40.30 g) = 0.01588 mol
Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO
The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol
Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles
0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL
The correct option is C.
The pitch of a string refers to the quality of sound that is produced by the string when it vibrates. There are three basic factors that affect the quality of pitch of a string, these are: the tension, the thickness of the string and the length of the string. The higher the thickness of the string, the lower the pitch of the string, thus, increasing the thickness of the string will lower the pitch.
