Answer:
The length of foil will be 8107.81 cm or 81.7081 m.
Explanation:
Given data:
Width of roll of foil = 302 mm
Height or thickness = 0.018 mm
Density of foil = 2.7 g/cm³
Mass of foil = 1.19 Kg
Length of foil = ?
Solution:
d = m/ v
v = length (l) × width (w) × height (h)
First of we will convert the Kg into gram and mm into cm.
one Kg = 1000 g
1.19 × 1000 = 1190 g
one cm = 10 mm
302 / 10 = 30.2 cm
0.018 / 10 = 0.0018 cm
Now we will put the values in formula:
d = m/ l× h× w
l = m / d × h× w
l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm
l = 1190 g/ 0.146772 g/cm
l = 8107.81 cm or 81.7081 m
PH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg[OH⁻]
pH=14+lg(8.7*10⁻¹²)=14-11.06=2,94
Lithium, sodium, potassium, rubidium, cesium, and francium
The final concentration of the diluted standard is 0.2 mg/dL.
<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>
Using the dilution formula:
where
- C1 is initial concentration
- V1 initial volume
- C2 is final concentration
- V2 is final volume.
Assuming a final volume of 100 mL, and since a 1/5 dilution is made:
C1 = 1.00 mg/dL
V1 = 20
C2 = ?
V2 = 100 mL
C2 = C1V1/V2
C2 = 20 × 1/100
C2 = 0.2 mg/dL
Therefore, the final concentration of the diluted standard is 0.2 mg/dL.
Learn more about dilution at: brainly.com/question/24881505
