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2 years ago

If there were only 4 protons, but the number of electrons remained at 6, what would the charge of this atom be? please help!!

Chemistry

1 answer:

lina2011 [118]2 years ago

5 0

Answer:

Negatively charged

Explanation:

It could be an ion too. To balance the electron shell it could lose 1 or 2 electrons. If it is not an ion it won't lose atoms to balance the shells.

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago

Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter

maksim [4K]

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(A) The given balanced reaction is,

$H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

$S_8(s)+8O_2(g)\rightarrow 8SO_2(g)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

$3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)$

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

$Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

$PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

6 0

3 years ago

Using ethanol and sodium or potassium cyanide as the sources of the carbon atoms, along with any necessary inorganic reagents, s

11Alexandr11 [23.1K]

Answer:

Please see attachment

Explanation:

Please see attachment

3 0

3 years ago

The principal source of sulfur on earth is deposits of free sulfur occurring mainly in volcanically Active regions. The offer wa

aivan3 [116]

Answer:

1.18x10⁸L of SO₂ and 2.36x10⁸L of H₂S

Explanation:

The balanced reaction is:

8SO₂(g) + 16H₂S(g) → 16H₂O(l) + 3S₈(s)

To solve this question we must find the moles of S₈ in 4.50x10⁵kg. With these moles and the reaction we can find the moles of SO₂ needed to react (Twice these moles = Moles Of H₂S needed). Using PV = nRT we can find the volume of the gas required:

<em>Moles S₈ - molar mass: 256.52g/mol-</em>

4.50x10⁵kg = 4.50x10⁸g * (1mol / 256.52g) =

1.75x10⁶ moles S₈

<em>Moles SO₂:</em>

1.75x10⁶ moles S₈ * (8mol SO₂ / 3mol S₈) = 4.68x10⁶ moles SO₂

<em>Moles H₂S:</em>

4.68x10⁶ moles SO₂ * 2 = 9.36x10⁶ moles H₂S

The volume could be obtained as follows:

PV = nRT

V = nRT / P

<em>V is volume in liters</em>

<em>n are moles: 4.68x10⁶ moles SO₂ and 9.36x10⁶ moles H₂S</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 22°C + 273.15 = 295.15K</em>

<em>P is pressure = 0.961atm</em>

<em />

Replacing:

Volume SO₂ and H₂S:

4.68x10⁶ moles * 0.082atmL/molK * 295.15K / 0.961atm =

<em>9.36x10⁶ moles H₂S</em> * 0.082atmL/molK * 295.15K / 0.961atm =

5 0

3 years ago

Because of the mud in the muddy water eventually settles out the muddy water is a ____________.

sp2606 [1]

it should be B. solution; a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).

witch basically means it sorts it self out when left a lone, like soap and water when not touched or the oil in the dressings you get at the store. hope this helped :D

6 0

4 years ago

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