Answer:
10.051kg of gold is the mass in kilograms
Explanation:
The density of gold is 19.3g/mL. To solve this question we need to convert the volume of the block of gold to grams using its density. Then, we must convert grams to kilograms using the equivalence factor (1000g = 1kg):
<em>Mass in grams:</em>
520.79mL * (19.3g / mL) = 10051g of gold
<em>Mass in kilograms:</em>
10051g of gold * (1kg / 1000g) = 10.051kg of gold
Answer:
Average atomic mass = 55.83 amu
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
For first isotope, Fe-54:
% = 6 %
Mass = 53.94 amu
For second isotope, Fe-56:
% = 92 %
Mass = 55.93 amu
For third isotope, Fe-57:
% = 2 %
Mass = 56.94 amu
Thus,
<u>Average atomic mass = 55.83 amu</u>
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
The highest point = crest
lowest point = trough
Answer:
This metal could be the aluminium with a specific heat of 
Explanation:
A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

Finally, the specific heat of this metal is

The aluminium could be the metal, its specific heat is similar to that found in this problem.
Finally, we can conclude that this metal could be the aluminium with a specific heat of 