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  • Andrea Apple opened Apple Photography on January 1 of the current year. During January, the following transactions occurred and were recorded in the books: 1. Andrea invested 13,500 cash in the business. 2. Andrea contributed 20,000 of photography equipment to the business. 3. The company paid 2,100 cash for an insurance policy covering the next 24 months. 4. The company received 5,700 cash for services provided during January. 5. The company purchased 6,200 of office equipment on credit. 6. The company provided 2,750 of services to customers on account. 7. The company paid cash of 1,500 for monthly rent. 8. The company paid 3,100 on the office equipment purchased in transaction #5 above. 9. Paid $275 cash for January utilities. Based on this information, the balance in the cash account at the end of January would be: Multiple Choice

Andrea Apple opened Apple Photography on January 1 of the current year. During January, the following transactions occurred and were recorded in the books: 1. Andrea invested 13,500 cash in the business. 2. Andrea contributed 20,000 of photography equipment to the business. 3. The company paid 2,100 cash for an insurance policy covering the next 24 months. 4. The company received 5,700 cash for services provided during January. 5. The company purchased 6,200 of office equipment on credit. 6. The company provided 2,750 of services to customers on account. 7. The company paid cash of 1,500 for monthly rent. 8. The company paid 3,100 on the office equipment purchased in transaction #5 above. 9. Paid $275 cash for January utilities. Based on this information, the balance in the cash account at the end of January would be: Multiple Choice

Andrea Apple opened Apple Photography on January 1 of the current year. During January, the following transactions occurred and were recorded in the books: 1. Andrea invested 13,500 cash in the business. 2. Andrea contributed 20,000 of photography equipment to the business. 3. The company paid 2,100 cash for an insurance policy covering the next 24 months. 4. The company received 5,700 cash for services provided during January. 5. The company purchased 6,200 of office equipment on credit. 6. The company provided 2,750 of services to customers on account. 7. The company paid cash of 1,500 for monthly rent. 8. The company paid 3,100 on the office equipment purchased in transaction #5 above. 9. Paid $275 cash for January utilities. Based on this information, the balance in the cash account at the end of January would be: Multiple Choice

Answer

  • Answer

Answer: $12,975

Explanation: To find the balance in the cash account at the end of January, we need to calculate the net cash flow by adding all cash inflows and subtracting all cash outflows.

Steps:

  1. Cash Inflows:
  • Andrea invested $13,500 cash.
  • The company received $5,700 for services provided.

Total Cash Inflows = $13,500 + $5,700 = $19,200

  1. Cash Outflows:
  • Paid $2,100 for an insurance policy.
  • Paid $1,500 for monthly rent.
  • Paid $3,100 on office equipment.
  • Paid $275 for January utilities.

Total Cash Outflows = $2,100 + $1,500 + $3,100 + $275 = $6,975

  1. Net Cash Flow:
  • Net Cash = Total Cash Inflows – Total Cash Outflows
  • Net Cash = $19,200 - $6,975 = $12,225

Therefore, the balance in the cash account at the end of January is $12,225.

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Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M−1 s−1) 1 OH + H2 → H + H2O 3.74 x 107 2 OH + HO2 → HO2 + OH− 5 x 109 3 OH + H2O2 → HO2 + H2O 3.8 x 107 4 OH + O2 → O2− + OH 9.96 x 109 5 OH + HO2 → O2 + H2O 7.1 x 109 6 OH + OH → H2O2 5.3 x 109 7 OH + eaq− → OH− 3 x 1010 8 OH + O2− → HO2 2.0 x 1010 9 H + O2 → HO2 2.0 x 1010 10 H + HO2 → H2O2 2.0 x 1010 11 H + H2O2 → OH + H2O 3.4 x 107 12 H + OH → H2O 1.4 x 1010 13 H + H → H2 7.9 x 109 14 eaq− + O2 → O2− 1.94 x 1010 15 eaq− + O2− → HO2− + OH− 1.3 x 1010 16 eaq− + HO2 → OH− + OH 2.5 x 1010 17 eaq− + H2O2 → OH + OH− 1.3 x 1010 18 eaq− + H → H− 2.5 x 1010 19 eaq− + eaq− + H2 + OH− 3.5 x 109 20 eaq− + H2O2 + OH− 4.5 x 109 21 HO2 + O2 → O2 + HO2 3.7 22 HO2 + HO2 → O2 + H2O2 3.7 23 HO2 + HO2 → O2 + OH + H2O 7 x 105 s−1 24 HO2− + O2 4.5 x 1010 25 H2O2 → 2OH 0.035 s−1 26 H+ + O2− → HO2 2 x 1010 27 H+ + HO2− → HO2 2 x 1010 28 H2O2 → H+ + HO2− 2.5 x 10−5 s−1 29 H2O2 → H+ + HO2− 2.5 x 10−5 s−1 30 O2− + O2 → HO2 + OH− 0.3 31 O2− + H2O2 → O2 + OH− + OH 16 32 1.8 x 107

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O → O + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ