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lana [24]
1 year ago
8

The following are electronegativity. What value should be where the M is?0.981.57M2.553.043.443.98No data

Chemistry
1 answer:
ipn [44]1 year ago
6 0

Answer 2.04

Explanation

Electronegativity decreases down the group and increases across the period in the periodic table.

In the period two of the periodic table,we have the following values for electronegativities with respect to its elements.

Li...........0.98

Be.........1.57

B..........2.04

C.........2.55

N..........3.04

O...........3.44

F...........3.98

Ne........n.a

The value that should be where M is is 2.04

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Which has a greater volume, 10 grams of water or 10 grams of acetone?
WINSTONCH [101]

Density is defined as the ratio of mass to the volume.

Density = \frac{Mass}{Volume}          (1)

Mass of water = 10 grams

Mass of acetone  = 10 grams

Density of water  = 1 \frac{g}{cm^{3}}

Density of acetone  = 0.7857 \frac{g}{cm^{3}}

Put the value of density of water and its mass in equation (1)

1 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

Volume of water =  10 cm^{3}

Put the value of density of acetone and its mass in equation (1)

0.7857 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

Volume of acetone = 12.72 cm^{3}

Thus, volume of acetone is more than volume of water because the density of acetone is lower.

6 0
3 years ago
In a reaction vessel of 10.0 I, heat 9.00 g of water so that it partially decomposes. In the equilibrium mixture, 0.250 mol of w
MatroZZZ [7]

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7 0
3 years ago
What is the temperature of 1.485 moles of N₂ gas at a pressure of 1.072 atm and a volume of 20 L?
morpeh [17]

Answer:

178.67K

Explanation:

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= 1.072atm*20L/1.485mol*0.0821LatmK^-1

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8 0
3 years ago
Will the solubility of CaSO4 in 0.10 Molar CaF2 be smaller than, larger than, or the same as in pure water?
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3 0
3 years ago
A 5.00 mL sample of hydrochloric acid is titrated with 0.1293 M ammonia (a base). If the titration required 28.15 mL of ammonia,
Sladkaya [172]

Answer:

1. C = 0.73 M.

2. pH = 0.14

       

Explanation:

The reaction is the following:

HCl + NH₃ ⇄ NH₄⁺Cl⁻

From the titration, we can find the number of moles of HCl that were neutralized by the ammonia.

n_{a} = n_{b}

Where "a" is for acid and "b" is for base.

The number of moles is:

n = C*V  

Where "C" is for concentration and "V" for volume.

C_{a}V_{a} = C_{b}V_{b}

C_{a} = \frac{0.1293 M*28.15 mL}{5.00 mL} = 0.73 M

Hence the initial concentration of the acid is 0.73 M.

The original pH of the acid is given by:

pH = -log([H^{+}])

pH = -log(0.73) = 0.14          

Therefore, the original pH of the acid is 0.14.

I hope it helps you!                  

4 0
3 years ago
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