The Density of the metal is 5.6 g/cm³
<h3>What is the density of a substance?</h3>
The density of a substance is the ratio of the mass and the volume of the substance.
The density of the metal is calculated as follows:
mass of metal = 1.4 kg = 1400 g
volume of metal = 3.2 * 17.1 * 4.6 = 251.712 cm³
Density of metal = 1400 g/251.712 cm³
Density of the metal = 5.6 g/cm³
Therefore, the density of the metal is obtained from the mass and the volume of the metal.
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The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.
<h3>What is the dissociation enthalpy?</h3>
Given that;
H-H bond energy = 435 kJ mol^-1
H-Cl bond energy = 431 kJ mol^-1
ΔHfO of HCL(g) = -92kJ mol^-1
Bond dissociation enthalpy of the Cl-Cl bond = x
-92 = 435 + 431 + x
x = -92 - (435 + 431)
x = -958 kJ mol^-1
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<span>Helium = 1
Carbon = 8
Nitrogen = 8
Strontium = 52
Tellurium = 71
If you look on a periodic table, on each element there is a number on
the top left. This represents the number of protons in an atom. Protons
have a mass of 1 (in relative to Carbon-13)
If we take nitrogen-15 for example; The number 15 tells you that the
isotope has a mass of 15. Now if you look on the periodic table,
Nitrogen has a proton number of 7. Only protons and neutrons have a
mass, electrons are considered to be negligable. Therefore the number of
neutrons Nitrogen-15 contains is 15 - 7 = 8 </span>
Answer:
2.11 g hydrobromic acid (correct to 3SF)
Explanation:
Molecular formula of hydrobromic acid = C2H5BrO2
mass of C2H5BrO2 = 140.96g
Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.
It looks like this:
9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2
Balanced Eqn
2
C
2
H
6
+
7
O
2
=
4
C
O
2
+
6
H
2
O
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume
60
⋅
300
224
=
80.36
g
ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =
4
⋅
44
⋅
80.36
60
=
235.72
g
and its no. of moles will be
235.72
44
=5.36 where 44 is the molar mass of Carbon dioxide
hope this helps
