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2 years ago

15

Which term means human-made liquid, solid, or sludge waste that may endanger human health or the environment? A. sewage B. hazar

dous waste C. groundwater D. non-biodegradable waste

2 answers:

zmey [24]2 years ago

4 0

The answer is B Hazardous waste.

iren2701 [21]2 years ago

4 0

B.) Hazardous waste.

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Which of these compounds would you expect to have the lowest melting point?

Leya [2.2K]

Answer:

Hmmm. #3

Explanation:

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2 years ago

The boiling points of halogens increase in the following order F2 a. ion-dipole<br> b. hydrogen-bonding<br> c. ion-ion<br> d. di

Viefleur [7K]

In order for the molecule to change phase from liquid to gas and evaporates, it needs to overcome the force from other molecules around it. as the force bigger evaporation gets harder. so e has the highest force and higher boiling point.

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2 years ago

Two substances with empirical formula HNO are hyponitrous acid (M = 62.04 g/mol) and nitroxyl (M = 31.02 g/mol).(c) Predict the

Levart [38]

While Nitroxyl, or HNO, has a dihedral form and just one nitrogen, Hyponitrous Acid, N2(OH)2, has both cis and trans structures and contains two nitrogens.

Hyponitrous acid has the chemical formula H2N2O2 or HON=NOH. 62.028 g/mol is the molecular weight of it. Additionally, it can take either a trans or cis form. When dry, the trans-hyponitrous acid crystallizes into white, explosive particles. Recent research has focused on nitroxyl (HNO), the one-electron reduced and protonated congener of nitric oxide (NO), as a potential drug for the treatment of heart failure and as a pre-conditioning agent to prevent ischemia-reperfusion injury, among other potential uses.

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5 0

1 year ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

<u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

<em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

<u>5. Calculate the number of moles of HgO</u>

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

<u>6. Convert to mass</u>

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

2 years ago