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2 years ago

What advice would you give a person who handles hydrogen and oxygen in their workplace?

1 answer:

3 0

Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.

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Write the formula for diphosphorous pentoxide

ehidna [41]

Answer:

P4O10

Explanation:

7 0

2 years ago

What's wrong with this thermochemical equation?

mafiozo [28]

This thermochemical equation needs to be balanced. Hence, option B is correct.

<h3>What is a balanced chemical equation?</h3>

A balanced equation contains the same number of each type of atom on both the left and right sides of the reaction arrow.

The balanced thermochemical equation is:

$H_2(g) + 2NaCl (s) + 3O_2 (g)$ → $2HCl (l) + 2Na (s) + 2O_3 (g)$

Hence, option B is correct.

Learn more about the balanced chemical equation here:

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7 0

1 year ago

What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?

zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

V2 = 500 mL

V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL

3 0

2 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

2 years ago

Draw the product(s) that are expected when tert-butyl bromide undergoes solvolysis in isopropanol, (CH3)2CHOH.

N76 [4]

Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo SN¹ and E¹ elimination reaction due to the formation of stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as nucleophile it will produce ether and when it acts as a base it will produce unsaturated compound. The reaction along with both products is shown below,

5 0

2 years ago

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