Answer:
it's urine option ( C ) .
Answer:
32.7 g of Zn
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Zn + 2HCl —> ZnCl₂ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂
Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.
Thus, 0.5 mole of Zn is required.
Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:
Mole of Zn = 0.5 mole
Molar mass of Zn = 65.4 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.5 × 65.4
Mass of Zn = 32.7 g
Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.
The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M
calculation
molarity =moles of solute/ Kg of the solvent
mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol
moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles
calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol
mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg
molarity is therefore = 0.06/0.09842 = 0.6 M
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>
Answer:
449730.879 cal/g
Explanation:
Given data:
Mass of sample = 4.9 g
Change in temperature = 2.08 °C (275.23 k)
Heat capacity of calorimeter = 33.50 KJ . K⁻¹
Solution:
C(candy) = Q/m
Q = C (calorimeter) × ΔT
C(candy) = C (calorimeter) × ΔT / m
C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g
C(candy) = 9220.205 KJ / 4.90 g
C(candy) = 1881.674 KJ / g
It is known that,
1 KJ /g = 239.006 cal/g
1881.674 × 239.006 = 449730.879 cal/g
