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3 years ago

Which element (Na or Cl) is more likely to steal an outer electron from the other?

Chemistry

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

Cl is more likely to than Na

Explanation:

The question asks for the element more take up an electron from the other

The ability of an atom to take take or give electrons in a chemical reaction, depends on its electron affinity and ionization energy, respectively

Chlorine, Cl, has the highest electron affinity in the periodic table which makes it attract electrons more towards itself

Sodium, Na, has a low ionization energy (about 2/5 of that of chlorine) than chlorine, Cl, making sodium more readily able to give its valence electron in a reaction

Therefore, Cl is more likely to take up an outer electron from Na.

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When bar-headed geese fly at very high altitudes (possibly over Mount Everest!), they breathe very thin air where the partial pr

Reil [10]

Answer:hemoglobin that has a high affinity for oxygen

Explanation:

Haemoglobin is the oxygen carrying pigment in blood. It performs this function because of the presence of iron at the center of the haemoglobin which coordinates reversibly with oxygen thereby aiding delivery of oxygen to cells. At high altitudes where air is thinner and the partial pressure of oxygen is lower than sea level, haemoglobin must develop a greater affinity for oxygen in order to carry the scarce oxygen to cells.

5 0

3 years ago

How many grams are in 2.3 moles of calcium phosphate, Ca3(PO3)2

irina1246 [14]

Moles = mass/molar mass
moles = 2.3
molar mass = 278
=> mass = moles*molar mass = 639.4g

6 0

3 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

Which part of an investigation is only found in an experimental investigation?

pantera1 [17]

Answer:

The correct answer is control group.

Explanation:

A group used in a study or in an experiment, which does not get treatment by the scientists and is used as a foundation to determine the functions of the other tested subjects is known as the control group. The control group is only found in an experimental investigation.

The group in an experiment, which gets the variable being examined is known as an experimental group. The comparison of an experimental group is done with a control group in order to find the answers in an experiment.

3 0

2 years ago

A ball is rolling across the floor. Every 3 seconds it travels 12 meters. Its speed, therefore, is 36 m/s. Is this problem solve

kirill [66]

No because you are supposed to do this

$d = t)v$
T for time and V for velocity or acceleration so what you do is take the two numbers

D= 3/12 it goes 12 meters every 3 seconds so you devide 12/3
$12 \div 3 = 4$
So for every second the ball roles 4 meters!
Hope this helps! :)

7 0

3 years ago