Answer is 3Mg + N2 —--> Mg3 N2
True.
A mixture is composed of two or more pure substance that are physically combined but are not combined chemically and thus no electrons (of their atoms) are involved.
The primary properties of a mixture include:
1. The components of a mixture are easily separated
2. The components each keep their original properties.
3. The proportion of the components may vary.
There are two main categories of mixtures:
1. Heterogeneous mixtures - substances are not evenly distributed e.g. oil and water mixture
2. Homogeneous mixtures - substances are evenly distributed throughout the mixture e.g. salt water, air.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
brainly.com/question/15289741
#SPJ1
Answer:
The correct option is;
c. The scientist can make and record observations
Explanation:
Empirical evidence are evidences obtained by direct observation, sensual perception or direct measurement. It is the processed and useful data gathered and stored in a material form or documented to provide record of the measurement.
With the aid of empirical evidence, it is possible for researchers to find answers to question regarding topics that can vastly impact every day life.
With empirical evidence, it is possible to determine the effects of treatment in a study such as the invention of working car safety devices such as seat belts and airbags.
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>
Therefore;
Moles of the compound will be;
= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
