Copper has a density of 8.96 g/cm^3
Answer:
31.9724 mL
Explanation:
The given antacid, Tumsr contains 40% of by mass.
Given mass of Tusmr = 400 mg
Mass of =
Also, 1 mg = 0.001 g
Mass of in Tusmr = 0.16 g
Molar mass of = 100.0869 g/mol
Moles = Mass / Molar mass = 0.16 g / 100.0869 g/mol = 15.9862 × 10⁻⁴ moles
Considering the reaction:
1 mole of react with 2 moles of HCl
15.9862 × 10⁻⁴ moles of react with 2*15.9862 × 10⁻⁴ moles of HCl
Moles of HCl = 31.9724 × 10⁻⁴ moles
Considering:
Or,
Given :
For HCl :
Molarity = 0.100 M
Let, Volume = x mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = x ×10⁻³ L
Moles of HCl = 31.9724 × 10⁻⁴
Volume = x = 31.9724 mL
Answer:
They would produce a repulsive force to another
Explanation:
A positive particle approaching another positive particle will repulse it.
According to coulomb's law "like charges repel one another and unlike charges attract".
A charge is an intrinsic property of any matter.
When like charges e.g positive and positive or negative and negative charges are in the vicinity of one another, they repel each other.
When unlike charges; positive and negative are brought together, they simply attract one another.
Therefore, we expect that a positive particle approaching another positive particle will repel one another.
2.8 g of CaCl2 is needed to react with 2.50 g of sodium carbonate.
<u>Explanation</u>:
- The balanced equation is given as
Na2CO3 +CaCl2 ----> 2 Nacl + CaCO3
- Use the balanced equation to determine the mole ratios between CaCO
3 and CaCl
2
.
Calcium carbonate and calcium chloride.
- Determine the moles of each reactant by dividing the given masses by their molar masses.
2.50 g of CaCO3 (1 mol of CaCO3 / 100.1 g of CaCO3) = 0.025 g of CaCO3.
- Determine the mass of CaCl
2 produced by each reactant by multiplying the moles of each reactant times the mole ratios with CaCl
2 in the numerator. Then multiply the result by the molar mass of CaCl
2 (
111 g/mol
)
.
0.025 g of CaCO3 (1 mol of CaCl2 / 1 mol of CaCO3) (111 g of CaCl2 / 1 mol of CaCl2) = 2.8 g of CaCl2.