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tresset_1 [31]
3 years ago
14

An element's atomic number is 62. How many electrons would an atom of this element have?

Chemistry
1 answer:
mylen [45]3 years ago
8 0

it has 62 electrons because science


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50.0 grams of KCl is dissolved in water to make a 4.00 L
xeze [42]

Answer:0.1677M

Explanation:

Molarity=moles/volume

Number of moles =mass/molar mass

Once you get the number of moles, you apply it to the molarity formula.

8 0
3 years ago
plz, help me Lisa’s doctor tells her that she needs to eat more foods that contain calcium to maintain her bone health. What cou
andriy [413]

Answer:

Deficiency of calcium which means that her body will take calcium from her bones which results in weakening of bones. (osteoporosis)

Explanation:

6 0
3 years ago
Sheena has a lump of sodium, a lump of potassium and a lump of lithium, but they’ve got mixed up and she doesn’t know which one
Arada [10]

Answer:

Explanation:

Just saw your request regarding answering this so here it is:

All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:

Most reactive: Potassium (K)

Intermediate: Sodium (Na)

Least reactive: Lithium (Li)

Hope it helps!

5 0
3 years ago
If you are told that one of them is Cr2O3 and the other is OsO4, which one do you expect to be the yellow solid?
Katen [24]
From own experience, I know that OsO4 is colorless, but Cr2O3 is actually a solid green from what I have seen. but I guess it could appear yellow-greenish.
7 0
4 years ago
Read 2 more answers
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

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

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

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

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

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

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

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

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

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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