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3 years ago

Ions Types of Ions Cation Anion 1. H+ 2.O-2 3.Ba2+ 4.Br- 5.Hg2+ 6.Li+ 7.Ca2+ 8.F- 9.S-2 10.Fe3+

Chemistry

1 answer:

fiasKO [112]3 years ago

6 0

Explanation:

The species with a positive charge over it, which means lack of electrons are called cations. Whereas the species with negative over it, meaning it has an excess of electrons with it are called anions.

From the list given the cations are

H+, Ba2+, Hg2+, Li+, Ca2+ and Fe2+

Whereas anions from the list are

O-2, Br-5, F-, S-2.

A Combination of cation and anions make an ionic compound in which cation is written first and then anion.

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Answer:

The candy is heated then cooled, turning solid like a igneous rock.

Explanation:

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3 years ago

Calculate the pH and pOH of 0.010 mol OH- per liter

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We can use a variety of formulas to determine our answers here.

Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid.

So, the pOH of OH- is 2.

To find pH we can set up this simple equation:
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3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

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Answer:

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