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Softa [21]
3 years ago
12

Why are there groups, in the periodic table?

Chemistry
2 answers:
Rasek [7]3 years ago
7 0
Groups help organize the elements. Elements of the same group have similar chemical qualities. So looking at an element’s group on the periodic table can help you identify some of it’s characteristics such it’s as reactivity with other elements or of it’s a good conductor.
zepelin [54]3 years ago
6 0
To classify the elements and make them easier to understand
You might be interested in
You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.
marishachu [46]

Answer:

Solution is 4.67% by mass of salt

Explanation:

% by mass is the concentration that defines the mass of solute in 100g of solution.

In this case we have to find out the mass of solution with the data given:

Mass of solution = Mass of solute + Mass of solvent

Solute:  Salt → 14.2 g

Solvent: Water → 290 g

Solution's mass = 14.2 g + 290g = 304.2 g

% by mass = (mass of solute / mass of solution) . 100

(14.2 g / 304.2g) . 100 = 4.67 %

3 0
3 years ago
Use Hess's Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given: 2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ
emmasim [6.3K]

Answer:

The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So,  to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ

Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ  

The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.

2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

2 ClF + 2 F₂ → 2 ClF₃

Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:

ClF + F₂ → ClF₃

Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>

3 0
3 years ago
Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga
vampirchik [111]
I guess a momogeneous mixture
7 0
3 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
What happens if the same zero error occurs in every reading
nadya68 [22]

Zero error occur in every reading means that the measuring instrument needs to recalibrated and adjusted.

<h3>What is Zero error?</h3>

This error occurs when a measuring instrument reflects a digit which isn't zero despite the real value being zero.

When this occurs, only the zero screw on the device should be adjusted so as to correct this technical error. This is why recalibration should be the most appropriate solution.

Read more about Zero error here brainly.com/question/4704005

#SPJ1

3 0
1 year ago
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