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Mrac [35]
3 years ago
11

Oxygen and sulfur are both in the family

Chemistry
1 answer:
Sav [38]3 years ago
8 0
On the periodic table, sulfur belongs to group 16, which is the chalcogen group or the oxygen family. This family contains oxygen, sulfur, selenium, tellurium and polonium.
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When light energy excites electrons in photosystem II, where do the electrons to replace them come from?
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hope this helped :)
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A piece of copper at 10 °C is dropped into a glass of water at 35 °C. Which statement is correct?
kakasveta [241]
I say B. Why? because the heat will keep on moving in molecules until it reaches same temperature
8 0
3 years ago
Read 2 more answers
Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G
dangina [55]

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change or heat required = ?

m = mass of water = 45 g

c_{p,s} = specific heat of solid water = 2.09J/g^oC

c_{p,l} = specific heat of liquid water = 4.18J/g^oC

c_{p,g} = specific heat of liquid water = 1.84J/g^oC

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

\Delta H_{vap} = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]

\Delta H=139287.75J=139.28775KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

3 0
3 years ago
Calculate the pH value in each of the following solutions, given their [H3O+] concentrations.
FrozenT [24]

Answer:

See Explanations ...

Explanation:

In general, pH is a 'p-factor' expression which, simply put, is a way to express very small numbers (i.e.; exponential data with 10⁻ⁿ value ranges) in a more convenient form. That is, by definition, pX = -log(X) where X is the data value of interest. In practical terms, p-factor analysis can be applied to a number of physical & chemical measurements such as ...

pH => measure of acidity of solution = -log[H₃O⁺]

pOH => measure of alkalinity of solution = -log[OH⁻]

pKa => measure of weak acid ionization in aqueous solution = -log(Ka)

pKb => measure of weak base ionization in aqueous solution = -log(Kb)

pKsp => measure of salt ionization in aqueous solution = -log(Ksp)

Such can be applied to ranges of small-number values defining other chemical and physical properties.

For this problem:

Gastric Juice: [H₃O⁺] = 1.6 x 10⁻²M => pH = -log(1.6 x 10⁻²) = -(-1.80) = 1.8

Cow's Milk:  [H₃O⁺] = 2.5 x 10⁻⁷M => pH = -log(2.5 x 10⁻⁷) = -(-6.60) = 6.60

Tomato Juice:  [H₃O⁺] = 5.0 x 10⁻⁵M => pH = -log(5.0 x 10⁻⁵) = -(-4.30) = 4.30

Other Applications:

Given:

[OH⁻] = 6.30 x 10⁻¹³M => pOH = -log(6.30 x 10⁻¹³) = -(-12.2) = 12.2

Ka = 4.5 x 10⁻⁵ => pKa = -log(4.5 x 10⁻⁵) = -(-4.35) = 4.35

Kb = 8.2 x 10⁻⁶ => pKb = -log(8.2 x 10⁻⁶) = -(-5.09) = 5.09

Ksp = 5.5 x 10⁻¹⁰ => pKsp = -log(-5.5 x 10⁻¹⁰) = -(-9.26) = 9.26

Note: The values for Ka, Kb & Ksp are typically provided in tables of weak acid ionization constants (Ka-values), weak base ionization constants (Kb-values) or solubility product constants of salts (Ksp-values).

Hope this helps, Doc :-)

4 0
3 years ago
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