How does the sun impact the speed of the water cycle?

The charges and sizes of the ions in an ionic compound affect the strength of the electrostatic interaction between the ions and

Citrus2011 [14]

Answer:

NaCl>MgCl2> MgS>KBr

Explanation:

The smaller the cation, the higher the lattice energy of the compound

8 0

3 years ago

Could I have help on 4 and 5? Just the letter please I don't need an explaintion

PIT_PIT [208]

5. is b
but im not sure about 4
hope this helps

6 0

3 years ago

rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o

SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

Initial important note:

Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.

1) Air:

Source: internet

Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.

2) Exhaled air:

Source: internet.

Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂

3) Air from the decomposition of H₂O₂.

In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:

i) Chemical Equation:

H₂O₂ (g) → H₂ (g) + O₂ (g)

ii) mole ratio of the products 1 mol H₂ : 1 mol O₂

iii) convert moles into mass (grams)

1 mol H₂ × 2 × 1.008 g/mol = 2.016 g

1 mol O₂ × 2 × 15.999 g/mol = 31.998 g

Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%

4) Air from the decomposition of NaHCO₃:

i) chemical equation:

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

ii) mole ratio: take into account only the gases in the products:

1 mol CO₂ (g) : 1 mol H₂O

iii) mass in grams

CO₂: molar mass ia approximately 44.01 g/mol

H₂O: molar mass is approximately 18.02 g/mol

iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.

5) The result is:

H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0

3 years ago

Read 2 more answers

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:

∵ ΔHsoln = Q/n

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

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ 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.

∴ final temperature = 25°C + 12.1 = 37.1°C.

6 0

2 years ago

The following three equations represent reactions done in this lab exercise. Classify each reaction by their type. Note there ma

gulaghasi [49]

Answer: 1. $Cu(s)+4HNO_3(aq)rightarrow Cu(NO_3)_2(aq)+2NO_2(g)+2H_2O(l)$ : oxidation reduction

2. $Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq)$ : precipitation

3. $CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l)$ : Double displacement

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$Cu(s)+4HNO_3(aq)rightarrow Cu(NO_3)_2(aq)+2NO_2(g)+2H_2O(l)$

Double displacement reaction is defined as the reaction where exchange of ions takes place. Double displacement reaction in which one of the product remain in solid form are represented by (s) after their chemical formulas. Such double displacement reaction are called as precipitation reaction.

$Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq)$

Double displacement reaction is defined as the reaction where exchange of ions takes place.

$CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l)$

Single displacement reaction is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.

Decomposition reaction is defined as the reaction where a single substance breaks down into two or more simpler substances.

Synthesis/Combination reaction is defined as the reaction where substances combine in their elemental state to form a single compound.

6 0

3 years ago