Answer:
The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g
Explanation:
We do this by preparing the equation:
Mass = concentration (mol/L) x volume (L) x Molar mass
Mass = 1.0 M x 2.5 L x 40 g/mol
Mass = 100 g
The specific gravity of the solution is 1.07, which contains 5.34g for a 5.00ml sample.
Density of a substance is
Density = Mass / Volume
= 5.34/5
= 1.07
Density of water = 1
Specific gravity = Density of substance/density of solvent
Specific gravity = 1.07 / 1
Specific gravity = 1.07
Specific gravity, often known as relative density, is the ratio of a substance's density to that of an industry standard.
If a material has a relative density that is less than 1 compared to the reference, it is less dense than the reference; if it is larger than 1, it is denser. Identical volumes of the two substances have the same mass if the relative densities are equal, or exactly 1.
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Answer:
Explanation:
Hello!
In this case, since the reaction between sulfuric acid and aluminum hydroxide is:
Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:
Then, given the molarity, it is possible to obtain the milliliters as follows:
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Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
The ratios which are needed to determine the mass of oxygen produced from the decomposition of 10 grams of potassium chlorate are;
- 31.998 g O2 : 1 mole O2
- 3 mole O2 : 2 mole KClO3
- 112.55 g KClO31 mole KClO3
From stoichiometry;
- We can conclude that according to the reaction;
3 moles of oxygen requires 2 moles of KClO3 to be produced.
And from molar mass analysis;
- 31.998 g O2 is equivalent to 1 mole O2
- O2112.55 g KClO3 is equivalent to 1 mole KClO3
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