Answer:
3.7mL is the volume of the object
Explanation:
To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:
7.9g * (1mL / 2.28mL) =
<h3>3.7mL is the volume of the object</h3>
The chemical formula for the compound containing 8.6 mol of sulfur and 3.42 mol of phosphorus is P₂S₅
<h3>How do I determine the formula of the compound?</h3>
From the question given above, the following data were obatined:
- Sulphur (S) = 8.6 moles
- Phosphorus (P) = 3.42 mole
- Chemical formula =?
The chemical formula of the compound can be obtained as follow:
Divide by their molar mass
S = 8.6 / 32 = 0.26875
P = 3.42 / 31 = 0.11032
Divide by the smallest
S = 0.26875 / 0.11032 = 2.44
P = 0.11032 / 0.11032 = 1
Multiply by 2 to express in whole number
S = 2.44 × 2 = 5
P = 1 × 2 = 2
Thus, the chemical formula is P₂S₅
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Answer:
Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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