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makkiz [27]
2 years ago
8

Need help with this.

Chemistry
2 answers:
Nostrana [21]2 years ago
8 0

Answer:

D

Explanation:

because it is changing to its own compound of chemicals.

ankoles [38]2 years ago
3 0
The answer would be D!
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Pls help I don’t understand at all
galina1969 [7]

Answer:

Mass = 12.48 g

Explanation:

Given data:

Mass of sulfur dioxide = 25.0 g

Mass of sulfur formed = ?

Solution:

Chemical equation:

SO₂      →        S   + O₂

Number of moles of SO₂:

Number of moles = mass/molar mass

Number of moles = 25.0 g / 64.07 g/mol

Number of moles = 0.39 mol

Now we will compare the moles of SO₂  with S.

                     SO₂        :          S

                        1          :           1

                       0.39     :         0.39

Mass of sulfur:

Mass = number of moles × molar mass

Mass = 0.39 mol × 32 g/mol

Mass = 12.48 g

4 0
2 years ago
recommend an element use to fill bottles that contain ancient paper. the element should be a gas at room temperature, should be
Eduardwww [97]

Okay so we are given these requirements:

element which can be used to stuff bottles that enclose ancient paper
must be a gas at room temperature
must be denser than helium
must not react with other elements

 

The only element that comes into my mind is:

<span>Argon</span>

6 0
3 years ago
For which of the following reactions is S° &gt; 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2
Lostsunrise [7]

Answer: 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

NH_4I(s)\rightarrow NH_3(g)+HI(g)

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

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. 2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

5 0
3 years ago
The strength of an atom's attraction for the electrons in a chemical bond is the atom's?
Mazyrski [523]

<span><span>Yes. An element that is highly electronegative pulls more on the electrons in a bond, such as oxygen in H20. This creates a polar bond, where there is a small negative charge on the oxygen, and a small positive charge in between the hydrogens.

</span>Credit goes to "Erin M" answered on yahoo answers a decade ago.
</span>
3 0
3 years ago
In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is
V125BC [204]

Answer:

the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

Explanation:

Given that:

mass of diphenylacetylene (C_{14}H_{10}) = 0.5297 g

Molar Mass of diphenylacetylene (C_{14}H_{10}) = 178.21 g/mol

Then number of moles of diphenylacetylene (C_{14}H_{10})  = \frac{mass}{molar \ mass}

= \frac{0.5297  \ g }{178.24 \  g/mol}

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = Heat absorbed by H_2O + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene (C_{14}H_{10}) will be = \frac{20598.87 \ J}{0.002972 \ mol}

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

3 0
3 years ago
Read 2 more answers
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