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VashaNatasha [74]

3 years ago

14

Which notation represents the largest atomic radius?

1 answer:

Darina [25.2K]3 years ago

7 0

Answer:

Calcium(Ca)

Explanation:

Atomic radius is the distance from the nucleus of an atom the outermost energy shell or energy level.
The atomic radii of metals are greater than the ionic radii since metals form ions by losing electrons.
Calcium atom has a larger atomic radius compared to magnesium atom because it has more energy levels. Calcium atoms have four energy levels while magnesium atoms have three energy levels.

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Protons + Neutrons = the atomic ?

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The atomic mass number is the number of protons and neutrons in an atom.

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Aluminum is an element. Which best describes what makes up a sample of aluminum?

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What are the four symbols for physical states of reactants and products?

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3 years ago

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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How many liters of 4M solution can be made using 100 grams of lithium bromide

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You can make 0.288 L of 4M solution using 100g of lithium bromide

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