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Nata [24]
3 years ago
6

1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the

Chemistry
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

Explanation:

Rubidium has a more negative reduction potential (-2.98 V) compared to strontium (-2.89 V).

Hence, in a redox reaction involving rubidium and strontium, rubidium will be oxidized while strontium is reduced.

The balanced redox reaction equation is obtained from;

Oxidation half equation;

2Rb(s) ---->2Rb^+(aq) + 2e

Reduction half equation;

Sr^2+(aq) + 2e ----> Sr(s)

Overall reaction equation;

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

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stepan [7]
Secondary growth is important to plants because it involves thickening of the plant axis.It also increased amounts of vascular tissue.

I tried sorry if it’s not worded perfect :)
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3 years ago
A bucket of water contains 5.45 L of water, how many mL of water is this ?
nalin [4]
Answer: It is 5450 mL


Explanation: There are 1000 mL in every L and then there is an extra 450 so just add that at the end
6 0
3 years ago
A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and
zlopas [31]

Answer:

37.1°C.

Explanation:

  • Firstly, we need to calculate the amount of heat (Q) released through this reaction:

<em>∵ ΔHsoln = Q/n</em>

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

  • We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>

6 0
3 years ago
A naturally occurring element consists of three isotopes. The data for the isotopes are:
max2010maxim [7]
(46.972* 69.472% + 48.961*21.667% + <span>49.954*8.8610%)/100% =

</span>= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u
7 0
3 years ago
How did Rutherford discredit Thomson's plum pudding model of an atom?
Dafna1 [17]

Answer:

He conducted an experiment using gold foil and alpha particles.

Explanation:

Ernest Rutherford in 1911 performed the gold foil experiment which provided a better outlook to the structure of the atom. In his experiment, he bombarded a thin gold foil with alpha particles. Most of the alpha particles passed through the gold foil and just a few was deflected back.

This observation led Rutherford to propose the nuclear model of the atom in which an atom has a small positively charged centre and electrons moving round it.

5 0
3 years ago
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