<span> Ethanol's </span>chemical<span> formula is C2H5OH.</span>Properties<span>. Pure ethanol is a flammable, colorless liquid with a boiling point of 78.5° C. Its low m</span>
The atomic mass of element B is 106.1 amu.
Given:
- The atomic mass of element A is 35.45 amu.
- A +B → AB
- 28.54 g sample of A combines with 85.42 g of another element B
To find:
The atomic mass of element B.
Solution:
The mass of element A = 28.54 g
The atomic mass of element A = 35.45 amu = 35.45 g/mol
Moles of element A :
= 
A +B → AB
According to reaction, one mole of A combines with 1 mole of B , then 0.8051 moles of A will combine with:

Moles of B element = 0.8051 mol
Mass of element B used = 85.42 g
The atomic mass of element B =M

The atomic mass of element B is 106.1 amu.
Learn more about moles of substance here:
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Answer:
1.104 J/g°C
Explanation:
Using Q = m × c × ∆T
Where;
m = mass of substance (g)
c = specific hear capacity (J/g°C)
∆T = change in temperature (°C)
For a colorimeter,
Q(water) = - Q(metal)
m. c. ∆T (water) = - m. c. ∆T (metal)
According to the information provided;
For water:
m = 28.0g
c = 4.184 J/g°C
∆T = (21.23 - 19.73°C)
For the metal:
m = 2.05g
c = ?
∆T = (21.23 - 98.88°C)
m. c. ∆T (water) = - m. c. ∆T (metal)
[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]
[117.152 × 1.5] = -[2.05 × c × (-77.65)]
175.728 = -[-159.1825c]
175.728 = 159.1825c
c = 175.728 ÷ 159.1825
c = 1.104
c = 1.104 J/g°C
Since we have the mass of the iron ingot we can get its volume by dividing the mass (4.578) by the density of iron (7874 kgm^-3) and we get that the volume is 5.8140716x10^-4 m^3. now we convert the volume from m^3 to cm^3(1m^3=1000000cm^3) and so by multiplying 5.8140716x10^-4 by <span>1000000 we get:581.4cm^3 (1 decimal place).
the volume for any cuboid is V=l x w x s we have the volume, the length and the width, so we plug in the values and solve for s
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the answer is rounded to three significant figures
hope i helped