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dimulka [17.4K]
3 years ago
15

Express 5601 in scientific notation

Chemistry
1 answer:
Tju [1.3M]3 years ago
8 0
The answer would be 5.60 x 10^3
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If you add a neutron to an atom how does it change ?
Kobotan [32]

Answer:

Its atomic mass increases by 1

An isotope of that element is fotmed with mass differences by 1

Explanation:

7 0
2 years ago
How do you test for carbon dioxide ?? Explain
g100num [7]

Answer:

Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate

Explanation:

. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white

5 0
2 years ago
Explain why the C―Ha bond is much more acidic than the C―Hb bond in pentan-2-one. Select the single best answer. 253 Ha is less
mafiozo [28]

Answer:

Ha is more acidic than Hb because loss of Ha forms a resonance-stabilized conjugate base.

Explanation:

The carbon atom that is next to the carbonyl group in pentan-2-one is known as the alpha carbon atom, this carbon atom bears the Ha, the alpha hydrogen atoms.

Ha is more acidic than Hb because, loss of Ha leads to the formation of a resonance stabilized enolate ion. This resonance stabilization of the ion formed makes loss of Ha an easier process than loss of Hb, hence the answer above.

3 0
3 years ago
The ionic radius for Na+ is 0.097 ηm and for Cl- is 0.181 ηm, the absolute value of the charge for each ion is 1.6x10-19 C, ε_o=
Vanyuwa [196]

Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

Explanation:

At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

Fr = -B/(r^n) but n = 9

Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

B = (2.953 × 10⁻⁹⁵) N.m⁹

6 0
3 years ago
How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of
Drupady [299]
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
5 0
3 years ago
Read 2 more answers
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