Use the References to access important values if needed for this question. It is often necessary...

Question

Question

Add a comment Improve this question Transcribed image text

Answer 1

Answer #1

(2.43 X106) (1.00X10-9) = 2.43 x 10-2 5.13 X 104 6.00x 10-4 = 8.558 107 (2.80X10-5) (9.45 X104) (3.10 x 105) (7.20x10-5) = 1.

Add a comment

Answer 2

Similar Homework Help Questions

Use the References to access important values if needed for this question. It is often necessary...

Use the References to access important values if needed for this question. It is often necessary to do calculations using scientific notation when working chemistry each of the following calculations. 1.00×10-8 + 5.70x10-5 – 2.13x104 + 9.07x105 1.19x104 2.56x106 (1.00x10-8)(5.70x10-5) Submit Answer Retry Entire Group 9 more group attempts remaining
Rererences Use the References to access important values if needed for this question. It is often...

Rererences Use the References to access important values if needed for this question. It is often necessary to do calculations using scientific notation when working chemistry problems. For practice, perfo 1.00 x 10- + 8.13 x 10 = 8.13 x 10-4 +1.00 x 10-8 1.25 x 10 6.45 x 10 (1.00 x 10-(8.13 x 10-4) Submit Answer Use the References to access important values It is often necessary to do calculations using scientific notation when working chemistryp (4.20 x 10-6)...
can you explain each steps please? [References Use the References to access important values if needed...

can you explain each steps please? [References Use the References to access important values if needed for this question. It is often necessary to do calculations using scientific notation when working chemistry problems. For practice, perform each of the following calculations. ) (1.00 x 10) (7.00 x 10- = 0.00000000000000 X 2.08 x 104 8.60 x 104 4) (7.14 x 10 (5.15 x 10)(2.08 x 10) 7.00 x 10-6 X Try Another Version 2 item attempts remaining Submit Answer Previous...

[References Use the References to access important valaes if needed for this It is often necessary...

[References Use the References to access important valaes if needed for this It is often necessary to do calculations using scientific notation when working chemist each of the following calculations. 6.13 x 10+4.66 x 10 4.66 x 104+1.00 x 10-8 7.71 x 104 4.57 x 10 o) (46 6.13 x 105 4.66 x 10
References Use the References to access important values if nes It is often necessary to do...

References Use the References to access important values if nes It is often necessary to do calculations using scientific notation when worki each of the following calculations. (1.64 x 100)(2.29 x 10) -[ 64 x 10° 3.38 x 104 2.40 x 10 1.00 x 10-3 (8.40 x 10 3.38 x 104 Il
Use the References to access important values if needed for this question. Taking logarihms and antilogarithms...

Use the References to access important values if needed for this question. Taking logarihms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following 53.7 0.296 0.936 Submit Answer Retry Entire Group 9 more group attempts remaining Previous Next

Review Topics) References Use the References to access important values if needed for this question. The...

Review Topics) References Use the References to access important values if needed for this question. The rate of effusion of SO2 gas through a porous barrier is observed to be 6.44 x 104 mol/h. Under the same conditions, the rate of effusion of NH3 gas would be mol/h. Submit Answer Retry Entire Group 9 more group attempts remaining
Use the References to access important values if needed for this question. The rate of effusion...

Use the References to access important values if needed for this question. The rate of effusion of SO2 gas through a porous barrier is observed to be 5.20 x 104 mol/h mol/h Under the same conditions, the rate of effusion of CO2 gas would be 8 more group attempts remaining Submit Answer Retry Entire Group
help! Use the References to access Important values if needed for this question. In the laboratory,...

help! Use the References to access Important values if needed for this question. In the laboratory, a general chemistry student measured the pH of a 0.575 M aqueous solution of quinoline, C.H.N to be 9.262. Use the information she obtained to determine the Kh for this base. Ky(experiment) - Submit Answer Retry Entire Group 8 more group attempts remaining In the laboratory, a general chemistry student measured the pH of a 0.319 M aqueous solution of diethylamine, (CH2)2NH to be...

Use the References to access important values if needed for this question. A buffer solution is...

Use the References to access important values if needed for this question. A buffer solution is made that is 0.372 M in HF and 0.372 M in KF. If K. for HF is 7.20 × 10-4, what is the pH of the buffer solution? 3l I pH = Write the net ionic equation for the reaction that occurs when 0.107 mol NaOH is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter.)...

Use the References to access important values if needed for this question. It is often necessary...

Homework Answers