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Answer:</h3>
The total concentration of ions in a 0.75 M solution of HCl is 1.5 M
That is; 0.75 M H⁺ and 0.75 M Cl⁻
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Explanation:</h3>
- Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
- The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
- For instance, HCl dissociates to give H⁺ and Cl⁻
HCl(aq) → H⁺(aq) + Cl⁻(aq)
- Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
- Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
Answer:
Due to gravitational Force the water exerts more pressure at "ground floor" than at "2nd floor".
Explanation:
Answer:
The empirical formula is, C4H4S
Explanation:
Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles
Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g
Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles
Mass of hydrogen = 0.025 moles × 1 = 0.025 g
Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles
Mass of sulphur= 0.0064 moles ×32 = 0.2 g
Now we obtain the mole ratios by dividing through by the lowest ratio.
C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles
C4H4S
KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which <span>then will react with 4-nitroaniline to form a diazonium salt.</span>
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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