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pashok25 [27]
3 years ago
15

How to round 0.000763 21cg

Chemistry
1 answer:
Alex_Xolod [135]3 years ago
6 0
0.000763 cg! You're welcome, and I hope this helped you
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1. Raw egg to boiled egg.
QveST [7]

Answer:

1.Irreversible

2.Physical

3.

6 0
3 years ago
Consider the mechanism. step 1:step 2:overall:2a⟶bb c⟶d2a c⟶dslowfast determine the rate law for the overall reaction, where the
Ne4ueva [31]

Answer:

Rate: R = k [A]²

Explanation:

The rate law equation for a given complex chemical reaction is given as the product of molar concentrations of the reactant raised to the power their respective partial reaction orders. The sum of the partial reaction orders is equal to the overall order of reaction.

For a complex reaction, the rate law is generally determined by the slowest step, which is known as the rate-determining step.

<u>Given reaction mechanism:</u>

Step 1: 2 A⟶ B,               slow step

Step 2: B+ C⟶ D,            fast step

Overall: 2 A + C ⟶ D

In this given reaction mechanism, the <em>step 1 is the slow step and thus the rate determining step.</em>

Therefore, the rate law for the given mechanism is:

Rate: R = k [A]²

Here, k is the overall rate constant

Also, the overall order of the reaction = 2

Therefore, the given chemical reaction is said to be second order reaction.

3 0
3 years ago
What is the cost of carbon in gram
Morgarella [4.7K]
Carbon is 12 grams per mole
6 0
2 years ago
An analytical chemist has determined by measurements that there are 96.5 moles of carbon in a sample of acetic acid. how many mo
Nadya [2.5K]
The formula of acetic acid is CH3COOH => C2H4O2.

So, the acetic acid has the same number of atoms of carbon (C) than of oxygen (O).

Therefore, the sample that contains 96.5 moles of carbon, will contain also 96.5 moles of O.

Answer: 96.5 moles of oxygen.
3 0
3 years ago
A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
5 0
3 years ago
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