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3 years ago

The boiling points of some group 7A hydrides are tabulated below.

Chemistry

1 answer:

11111nata11111 [884]3 years ago

6 0

hey there!:

In the case of ammonia, the hydrogen bond is formed using the lone pair present in nitrogen and the hydrogen having δ+ charge (due to bonding with electronegative N) of another ammonia molecule. Thus the inter molecular attraction increases which in turn increases the boiling point .

Option B is the correct answer.

Hope this helps!

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What is the number of copper atom in one naira coin which weight is 7.3g a sum of the material is 86%

Step2247 [10]

Answer:

5.9x10²² atoms of Cu are in one naira coin

Explanation:

To solve this question we need to find the mass of Copper. Then, using its molar mass (Cu = 63.546g/mol) we must find the moles of Cu and its atoms using Avogadro's number:

Mass Cu:

7.3g * 86% = 6.278g is the mass of Cu.

Moles Cu:

6.278g * (1mol / 63.546g) = 0.099moles Cu

Atoms Cu:

0.099moles Cu * (6.022x10²³atoms / 1mol) =

<h3>5.9x10²² atoms of Cu are in one naira coin</h3>

4 0

2 years ago

If a bullet travels at 407.0 m/s, what is its speed in miles per hour? Number click to edit mi/h For the same bullet travelling

kaheart [24]

Answer : The speed in miles per hour is 22 mile/hr.

The speed in yard per min is 26617.8 yard/min

Explanation :

The conversion used for meters to miles is:

$1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}$

The conversion used for second to hour is:

$1s=\frac{1}{60}min\times \frac{1hr}{60min}$

The conversion used for meter per second to mile per hour is:

$1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in miles per hour.

$1m/s=2.2mile/hr$

So, $407.0m/s=\frac{2.2mile/hr}{1m/s}\times 407.0m/s=895.4mile/hr$

Therefore, the speed in miles per hour is 22 mile/hr.

The conversion used for meter to yard

1m = 1.09 yard

The conversion used for second to hour is:

$1s=\frac{1}{60}min$

The conversion used for meter per second to mile per hour is:

$1m/s=\frac{1.09yard}{frac{1}{60min}}$

$1m/s=65.4yard/min$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in yards per min

$1m/s=65.4yard/min$

So, $407.0m/s=\frac{65.4yard/min}{1m/s}\times 407.0m/s=26617.8yard/min$

Therefore, the speed in yard per min is 26617.8 yard/min

7 0

2 years ago

How are elements and compounds related

Lisa [10]

Elements are a one of a class of substances that cannot be seperwted into simpler substances by chemicalk means. Compounds is a substance formed when two or more chemical elements are chemically bonded together. Relation: When two elements or more are formed chemically together its called a compound without elements there won't be compounds and without compounds there won't be elements.

5 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

Can some right a paragraph about photosynthesis pls =))))))))))))))))))))))

Alina [70]

Answer:

Photosynthesis is the process in which plants harness energy from the sun to make their own nutrition. The process converts light energy into chemical energy through cellular respiration.

Explanation:

Just to start you off. I hope it helps! (❛ ᴗ ❛)

3 0

2 years ago

Read 2 more answers