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3 years ago

11

Group 1A element sodium combines with Group 7A element chlorine to form sodium chloride or table salt. Based on this information

, which expression is true about the electronegativity values of these two elements?
A) Na > Ci
B) Na < Ci
C) Na = Ci
D) We cannot make a prediction with data provided.

2 answers:

EastWind [94]3 years ago

8 0

A.<span>Na > Cl
is the right answer.</span>

nlexa [21]3 years ago

4 0

The correct answer in this question is option B. The electronegativity increases across a group. Chlorine ion has a higher electronegativity than the sodium ion. Chlorine need one electron to have a noble gas configuration that is why it gets one electron every chance it makes.

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A 4.24 kg marble slab has the volume of 1564 cm3 What is the density in g/cm3

miskamm [114]

Answer:

The answer is

<h2>2.71 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

volume of marble = 1564 cm³

1 kg = 1000 g

4.24 kg = 4240 g

mass = 4240 g

The density is

$density = \frac{4240}{1564} \\ = 2.710997442...$

We have the final answer as

<h3>2.71 g/cm³</h3>

Hope this helps you

7 0

2 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago

A substance that can be broken down into simpler subtend only by a chemical change is aa. homogenous mixtureb. heterogeneous mix

Natasha2012 [34]

Answer:

d. compound

Explanation:

<u>Mixture </u>: It is defined as the substance that is made by the combination of two or more different components which are physically combined.

<u>Homogeneous mixture</u> are the mixtures in which the composition of the substances in it are uniformly mixed. <u>Heterogeneous mixture</u> are the mixtures in which the composition of the substances in it are not uniformly mixed.

<u> Compound :</u> It is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass and are chemically bonded.

<u> Element :</u> It is a pure substance which is composed of atoms of similar elements.

<u>As discussed above, The different elements in the compounds are chemically bonded and hence, it is the substance which can be broken down into the simpler units when it subtend only by the chemical change.</u>

3 0

3 years ago

7. If a waste container has 87.0 mL of 0.234 M acetic acid, how many milliliters of 2.00 M sodium

Naya [18.7K]

Answer:

$V_{base}=10.2mL$

Explanation:

Hello.

In this case, since the neutralization of the acid requires equal number of moles of both acid and base:

$n_{acid}=n_{base}$

Whereas we can express it in terms of concentrations and volumes:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we can compute the volume of sodium hydroxide (base) as follows:

$V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{87.0mL*0.234M}{2.00M}\\ \\V_{base}=10.2mL$

Best regards.

8 0

2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

2 years ago