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2 years ago

The elements in Groups 1A(1) and 7A(17) are all quite reactive. What is a major difference between them?

Chemistry

1 answer:

matrenka [14]2 years ago

6 0

The elements in Groups 1A(1) and 7A(17) are all quite reactive.

<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>

Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.

Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.

Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.

To know more about Groups 1A(1) and 7A(17) please click here :

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If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of wate

lapo4ka [179]

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

<h3>Combustion:</h3>

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

$Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J$

<h3>Calorimetry:</h3>

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

$Q=mC\Delta T$

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

$m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g$

Finally, we convert it to liters as required:

$V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL} \\\\V=42.27L$

Learn more about calorimetry: brainly.com/question/1407669

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2 years ago

A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus

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Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>

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4 years ago

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