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Arlecino [84]
3 years ago
7

PLEASEEEEE ANSWERRR??

Chemistry
1 answer:
alekssr [168]3 years ago
3 0
I believe it is Good
Please give me Brainliest
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The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.
denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium (Be) belongs to the second group of the periodic table. The ground state electronic configuration of Be is:1s^{2}2s^{2}

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (2s^{2}).

After losing all valence electrons that is 2 electrons from 2s orbital. The electronic configuration will be:

1s^{2}2s^{0}

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as Be^{2+}.

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0
3 years ago
Read 2 more answers
Find the mass of 4.5 moles of H3PO4
Contact [7]

Hey there!

H₃PO₄

Find molar mass.

H: 3 x 1.008 = 3.024

P: 1 x 30.97 = 30.97

O: 4 x 16 = 64

---------------------------------

                  97.994 grams

The mass of 1 mole of H₃PO₄ is 97.994 grams.

We have 4.5 moles.

97.994 x 4.5 = 440

The mass of 4.5 moles of H₃PO₄ is 440 grams.

Hope this helps!

6 0
3 years ago
At a given temperature, 300m of carbon dioxide has a mass of 5.94kg. What is the density of carbon dioxide at this temperature
erica [24]
I think the would have to be 0.198kg/g
4 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f
Yanka [14]

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

#SPJ4

6 0
1 year ago
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