Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.
<h3>What are ions?</h3>
Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.
We have to calculate the total number of ions in 38.1 g of .
The molar mass of SrF₂ = 125.62 g/mol
The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles
Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles
Total moles of ions in 0.30329 moles of
= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions
We know that,
1.0 mole of ions = 6.023 x 10²³ ions
Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions
Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions
To learn more about ions, refer to the link:
brainly.com/question/14295820
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Answer:
26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃
Explanation:
To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?
moles of O₂= 26.25 ≅ 26.3
<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>
Answer:
S/.486 es el valor del anillo
Explanation:
Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.
Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:
90g × 59.1% = 53.19g Oro en el anillo
Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:
53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.
Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:
0.27mol × (S/.1800 / 1mol oro) =
<h3>S/.486 es el valor del anillo</h3>
Carbon atom with iron and helium
