All categories

3 years ago

0.00099989 3 sig figs, scientific notation

Chemistry

1 answer:

mixas84 [53]3 years ago

5 0

1.00*10^3
You’d need to lower the exponent because rounding to 3 sig figs changes the 9’s to - 1000. Keep the 0’s.

You might be interested in

Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Ksivusya [100]

Answer:

A. K₂C₂O₄ Potassium oxalate

B. CuC₂O₄ Copper oxalate

C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate

D. Pb(C₂O₄)₂ Lead (IV) oxalate

E. (NH₄)₂C₂O₄ Ammonium oxalate

F. HC₂O₄⁻ Acid oxalate

Explanation:

C₂O₄⁻² → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓

This is a precipitate.

C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓

This is a precipitate.

D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓

This is a precipitate.

E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻

HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka

HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb

HC₂O₄⁻ is an amphoteric compound

6 0

3 years ago

What is the Na+ concentration in each of the following solutions:

iren [92.7K]

Sodium Sulfate = Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M stands for Molarity, defined as Molarity = (moles of solute)/(Liters of solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65 moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+ ions in solution for every one Na2SO4.

Therefore if 3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and since this is still a theoretical solution, we can assume 1 L of solution.

Finally we find [Na+] = 2*3.65 = 7.3M

Use the same logic for parts b and c

5 0

3 years ago

The following do not represent valid ground-state electron configurations for an atom either because they violate the Pauli excl

svetoff [14.1K]

Answer:

<em>For both cases the answer is C</em>

Explanation:

We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.

For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.

7 0

2 years ago

Read 2 more answers

Leslie wants to know which advertising medium works best to promote her electronics store. She surveys her customers over a week

Korolek [52]

No survey to base on

7 0

3 years ago

Read 2 more answers

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole