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2 years ago

Chlorine is much more apt to exist as an anion than is sodium. this is because __________.

1 answer:

s2008m [1.1K]2 years ago

6 0

The answer is that “chlorine has a greater electron affinity than sodium does”. Sodium has a lesser electron affinity than Chlorine, which is why it is deliberated that Chlorine is much more suitable to be existent as an anion than sodium, due to the lack of its electron affinity.

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A twenty-eight-liter volume of gas contains 11 g methane, 1.5-gram nitrogen and 16-gram carbon dioxide. Determine partial pressu

GREYUIT [131]

Based on Dalton's Law, for a mixture of gases, the total pressure is the sum of the partial pressure of each gas.

Partial pressure (p) of each gas is related to the total pressure (P) as follows:

p = X * P----------(1)

where X is the mole fraction of that gas

X = moles of a particular gas/total number of moles of all gases in the mixture--------------(2)

Step 1: Calculate the moles of each gas

Mass of methane, CH4 = 11 g

Mass of nitrogen, N2 = 1.4 g

Mass of carbon dioxide, CO2 = 16 g

# moles of CH4 = 11 g/16 gmol-1 = 0.6875

# moles of N2 = 1.4/28 = 0.05

# moles of CO2 = 16/44 = 0.3636

Total moles = 0.6875+0.05+0.3636 = 1.1011

Step2: Calculate mole fractions of each gas

Based on equation (2)

X(CH4) = 0.6875/1.1011 = 0.6244

X(N2) = 0.05/1.1011 = 0.0454

X(CO2) = 0.3636/1.1011 = 0.3302

Step 3: Calculate the total pressure

Based on ideal gas equation:

PV = nRT

given that;

V = 28 L

n = total moles = 1.1011

R = gas constant = 0.0821 Latm/mol-K

Since temp T is not given, let us consider room temperature of 25 C = 25 + 273 = 298 K

Now, P = nRT/V = 1.011*0.0821*298/28 = 0.962 atm

Step 3: Calculate partial pressures

Based on equation:

p(CH4) = 0.6244*0.962 atm = 0.601 atm

P(N2) = 0.0454*0.962 atm = 0.044 atm

P(CO2) = 0.3302*0.962 atm = 0.318 atm

4 0

3 years ago

In an effort to try winning World War I, many countries broke previously created treaties and pacts that banned the use of chemi

nadya68 [22]

D the belguim people atacked using it

3 0

3 years ago

Describe how to do a flame test on a sample of a salt

Katen [24]

Answer:Hope this helps!

Explanation:

You can use a flame test to help identify the composition of a sample. The test is used to identify metal ions (and certain other ions) based on the characteristic emission spectrum of the elements. The test is performed by dipping a wire or wooden splint into a sample solution or coating it with the powdered metal salt. The color of a gas flame is observed as the sample is heated. If a wooden splint is used, it's necessary to wave the sample through the flame to avoid setting the wood on fire. The color of the flame is compared against the flame colors known to be associated with the metals.

4 0

3 years ago

Which scientist arranged the elements in order of increasing atomic number, rather than increasing atomic mass?

pentagon [3]

Answer:

Henry Moseley

Explanation:

Dmitry Mendeleef and Lothar Meyer proposed a periodic table based on the atomic mass.

They stated a periodic law expressed as "chemical properties of elements are a periodic function of their atomic weights".

But, Henry Moseley in 1900s re-stated periodic law by changing the basis of the law from atomic weight to atomic number.

The present periodic law is stated as "the properties of elements are a periodic function of their atomic number".

6 0

2 years ago

The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6

s344n2d4d5 [400]

Answer:

the conversion factor is f= 6 mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0

3 years ago