Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
Answer:
5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)
Explanation:
mole conversions:
grams to moles => divide by formula wt.
moles to grams => multiply by formula wt.
gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)
This problem:
mass to moles => divide by formula wt.
mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)
When the temperature increases, the kinetic energy of the particles increase this inturn makes them vibrate
Answer:
Explanation:
Did you mean: V = d/t a = (V - Vit Average = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Showing results for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = "-9.81" m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Search instead for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
