<span>convert grams to moles then use the equatino to do moles of Na to moles of H2O then convert moles of H2O to molecules by using avogadros number (6.022e23)
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
1) The names of the molecules
2) Write the chemical equation
3) Balance the chemical equation
1 fluoride in the reactants and 1 fluoride in the products
1 carbon in the reactants and 1 carbon in the products
4 hydrogens in the reactants and four hydrogens in the products
1 oxygen in the reactants and 1 oxygen in the products
Since the number of atoms in the reactants is the same that the number of atoms in the products, the chemical equation is balanced.
Answer:
52.9 KJmol-1
Explanation:
From;
log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)
The temperatures must be converted to Kelvin;
T1 = 25° C + 273 = 298 K
T2= 35°C + 273 = 308 K
R= gas constant = 8.314 JK-1mol-1
Substituting values;
log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)
Ea = 52.9 KJmol-1
Answer:
The reactivity of metal is determined by the reactivity series. ... The metal which easily displaced aluminium will lie above in the series but that same element cannot displace sodium, so it will lie below in the series. Hence, from the series, we conclude that the unknown metal could be calcium or magnesium.
Explanation:
Hope this helps! :)
The answer is 57.14%.
First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.
Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g
Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g
mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%
Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
