Answer: 13 grams
Explanation:
The quantity of heat energy (Q) released from a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = 202.8 Joules
Mass of silver = ?
C = 0.240 J/g °C.
Φ = 65°C
Then, Q = MCΦ
202.8J = M x 0.240 J/g °C x 65°C
202.8J = M x 15.6 J/g
M = (202.8J / 15.6 J/g)
M = 13 g
Thus, the mass of silver is 13 grams
Answer:
V1 = 3.09 L
Explanation:
Initial Pressure, P1 = 15.6 psi
Initial Volume, V1 = ?
Final Pressure, P2 = 25.43 psi
Final Volume, V2 = 1.895 L
The relationship between these quantities is given by boyles law;
V1P1 = V2P2
V1 = V2P2 / P1 = 1.895 * 25.43 / 15.6
V1 = 3.09 L
Here we have to draw the mechanism of the reduction reaction between benzaldehyde and sodium borodeuteride to form the corresponding alcohol.
The reducing agent sodium borodeuteride can reduce the aldehydes to its corresponding alcohol. The reaction mechanism is shown in the attached image.
The reaction mechanism can be explained as-
The sodium borodeuteride is highly ionic in nature thus it remains as Na⁺ and BD₄⁻ The deuterium atom of BD₄⁻ attack the carbonyl carbon atom and substitute one of its deuterium as shown in the figure.
One molecule of sodium borodeuteride can reduce four molecules of benzaldehyde. The polar solvent like alcohol donates the proton as shown in the mechanism.
The converted alcohol contains the deuterium atom at the -C center. Thus benzaldehyde is converted to deuteroted benzyl alcohol.
The chemical equation represents the reaction describes is;
4NH3 + 5O2 = 4NO + 6H2O
Therefore 4 moles of NH3 reacts with 5 moles of O2.
1 mole of O2 (molar mass) = 2 * 16 = 32g.
5 moles of O2 = 5 * 32 = 160g
4 moles of NH3 = 4 (14 + 3*1) = 68g
Therefore, 68g of NH3 reacts with 160g of O2.
But, we have only 4.5 g of oxygen.
68g reacts with 160g
Xg reacts with 4.5
X = 68*4.5 / 160 = 1.9125g
The correct answer for the question that is being presented above is this one: "D. 15 millimeters." An object 5 millimeters high is located 15 millimeters in front of a plane mirror. T<span>he image of the mirror is located 15 millimeters.</span>
