2C4H10 + 13O2 = 8CO2 + 10H2O
1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10
2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O
3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O
<u>Answer:</u> The true statement is iron can reduce 
to gold metal
<u>Explanation:</u>
Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.
The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.
Metal A is more reactive than metal B.
We are given:
Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.
The increasing order of reactivity thus follows:
where, sodium is most reactive and gold is least reactive
For the given options:
<u>Option 1:</u> Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.
<u>Option 2:</u> Iron cant easily reduce gold ion to gold metal because it is more reactive.
<u>Option 3:</u> Silver cannot easily reduce iron ion to iron metal because it is less reactive.
Hence, the true statement is iron can reduce to gold metal
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
