Answer:
The correct options include choice 2, 3, and 6.
Explanation:
Density is identified as the mass of a substance per unit volume occupied by that substance.
The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.
2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.
3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.
6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.
The same metals in both instances will yield consistent densities due to the fixed density of the metal.
The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:
The substances 
and 
combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.
.
.
We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of 
equal 24.977 moles.
Next, we apply the ideal gas equation 
, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V
The final answer should be expressed in liters, 
, hence
Response: The moles in 369 grams of calcium hydroxide are 4.98 moles
Reasoning: Given,
Mass of calcium hydroxide = 369 g
Molar mass of calcium hydroxide = 74.093 g/mole
Formula used:
Now substituting the provided values into this formula, you will find the moles of calcium hydroxide.
Thus, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles
The interaction between calcium carbonate and hydrochloric acid can be represented by the chemical equation,
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
Calcium carbonate has a molecular weight of 100 g/mol, while hydrochloric acid's molecular weight is 36.45 g/mol. According to the equation, 100 g of calcium carbonate reacts with 72.9 g of hydrochloric acid.
x = (4 g HCl)(100 g CaCO3 / 72.9 HCl)
x = 5.49 g
Final result: 5.49 g
First, we need to identify the half-reaction for magnesium. It can be represented as:
Mg2+ + 2e- = Mg
Next, we will determine the overall charge generated during the electrolysis using the information derived from the half-reaction. The calculation follows:
4.50 kg Mg (1000 g / 1 kg) (1 mol / 24.305 g) (2 mol e- / 1 mol Mg) (96500 C / 1 mol e-) = 35733388.2 C
The provided EMF is given in voltage. Since 1 V equals J/C, 5 V translates to 5 J/C.
Therefore, 35733388.2 C (5 J/C) = 178666941 J
Finally, 178666941 J (1 kW-h / 3.6x10^6 J) = 49.63 kW-h
