density= mass/volume
we need the volume of the metal to find the density, in which case the answer would be 64.2/volume=density
Answer:
A = 0.75 ×10² KJ.
B = 3.9 ×10³ dg
C = 0.22 × 10² μl.
Explanation:
A = 7.5 ×10⁴ j to kilo joules
7.5 ×10⁴ / 1000 = 0.75 ×10² KJ.
Joule is the smaller unit while kilo joule is the larger unit. One kilo joule equals to the thousand joule that's why we will divide the given value by 1000 in order to convert into KJ.
B = 3.9 ×10⁵ mg to decigrams.
3.9 ×10⁵ / 100 = 3.9 ×10³ dg
Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into decigram we have to divide the value by 100.
C = 2.21 ×10⁻⁴ dL to micorliters
2.21 ×10⁻⁴ ×10⁵ = 0.22 × 10² μl.
Deciliter is bigger unit then micro liter . One deciliter equals to the 100000 micro liters. In order to convert the dL into micro liter we have to multiply the given value with 100000.
The most abundant carbon isotope is carbon-12.
The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.
the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.
To solve this we assume
that the gas inside is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
T2 = T1 x V2 / V1
T2 = 280 x 20.0 / 10
<span>T2 = 560 K</span>
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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