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saveliy_v [14]
3 years ago
14

Use the table to determine the charge of phosphate in K3PO4.

Chemistry
1 answer:
sveta [45]3 years ago
3 0

Answer:

                    Charge on Phosphate is -3 (PO₄³⁻)

Explanation:

                     Ions or polyatomic ions are formed when an atom or molecule loses or gains electron(s) respectively.

Example:

              Formation of ions.

                           Na  →  Na⁺  +  e⁻ (Na loses one electron)

                           Cl₂  +  2 e⁻  →  2 Cl⁻ (Cl gains electron)

In polyatomic ions a charge is present on a molecule which has either two or more atoms.

Example:

               CO₃²⁻ , SO₄⁻ , NH₄⁺ e.t.c.

In statement the  K₃PO₄ polyatomic ionic compound. It is made up of three K⁺ ions and one Phosphate ion (PO₄ˣ⁻) ion.

As we know K belongs to group 1 metals so it has one valence electron and thus, forms a cation of +1 charge. So, for a neutral polyatomic ionic compound the overall charge on K₃PO₄ should be zero. Hence,

                                         K (3) + PO₄ (X)  =  0

Or,

                                         +3 + X = 0

Or,

                                         X  =  0 - 3

                                         X  =  -3

Therefore, the charge on phosphate is -3.

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What change in volume results if 60 mL of a gas is cooled from 33 C to 5 C?
Eva8 [605]

Answer:

Change in volume on changing temperature from 33^{\circ}C to 5^{\circ}C is 5.49 mL

Explanation:

Initial volume of gas = V = 60 mL

Assuming final volume of gas to be V' mL

Initial temperature = T = 33^{\circ}C = 306 K

Final temperature = T' = 5^{\circ}C = 278 K

The relationship between volume and temperature of gas at constant pressure is shown below

\displaystyle \frac{V}{V'}=\displaystyle  \frac{T}{T'} \\\displaystyle \frac{60\textrm{ mL}}{V'} = \displaystyle \frac{306\textrm{ K}}{T} \\V' = 54.51 \textrm{ mL} \\\textrm{Change in volume} = \left ( V-V' \right ) \\\textrm{Change in volume} = \left ( 60-54.51 \right )\textrm{ mL} \\\textrm{Change in volume} = 5.49 \textrm{ mL}

Change in volume on changing temperature = 5.49 mL

6 0
3 years ago
Which picture accurately represents the type of orbital occupied by the electrons in a beryllium atom
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Answer:

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Explanation:

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6 0
2 years ago
Why would you need to heat the hydrate up to 250 degrees if water boils at 100 degrees Celsius? Do you think that the boiling po
sergeinik [125]

Explanation:

The pure form of water has a boiling point of 100°C. Boiling point is a physical property of matter and it shows that at such temperature, a liquid substance will change state to vapor.

Pure water is made up of 2 atoms of hydrogen and 1 atom of oxygen. The only intermolecular forces between them is the hydrogen bonds that must be broken for the water to boiling off.

In hydrate, water is present in another form. The water is attached to another compound.

For a pure liquid, the they have reasonably constant boiling point and low boiling range.

Impurities such as the other part of the hydrate causes the elevation of the boiling point and the widening of the boiling range for impure substances.

We are no longer dealing with just hydrogen bonds, other molecular interactions are now involved and they need to be accounted for.

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Pure substances brainly.com/question/1832352

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8 0
3 years ago
Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and
swat32

Answer:

\%\ mass\ of\ CaCO_3=93.37\ \%

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of CO_2 produced =  0.0493 moles

According to the reaction:-

CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of CaCO_3 = 100.0869 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0493\ mol= \frac{Mass}{100.0869\ g/mol}

Mass_{CaCO_3}=4.93\ g

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100

\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100

\%\ mass\ of\ CaCO_3=93.37\ \%

3 0
3 years ago
The density of water is 1 gm/ml. An object has a mass of 58 grams. What volume must it have in order to float in water?
Keith_Richards [23]

Answer:

58mL

Explanation:

Given parameters:

Density of water  = 1g/mL

Mass of object  = 58g

Unknown:

The volume the object must have to be able to float in water = ?

Solution:

To solve this problem, we know that the object must have density value equal to that of water or less than that of water to be able to float.

We then set its density to that of water;  

   Density  = \frac{mass}{volume}  

      Volume  = \frac{mass}{density}  

So;

      Volume  = \frac{58}{1}   = 58mL

7 0
2 years ago
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