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Stolb23 [73]
3 years ago
12

The answer to a chemistry question

Chemistry
1 answer:
Otrada [13]3 years ago
5 0
<span>4 C3H5(NO3)3 = 12 CO2 + 6 N2 + 10 H2O + O<span>2
That's the answer lovely~</span></span>
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A 135-g sample of a metal requires 2.50kj to change its temperature from 19.5 degrees Celsius to 100.0 degrees Celsius. What is
PolarNik [594]
Q = m x c x ΔT

2500 = 0.135 x C x 80.5

2500 = 10.8765 x C

C = 230.043 J/Kg.K

hope this helps
7 0
3 years ago
Read 2 more answers
Cl2 + 2OH− → Cl− + ClO− + H2O
Nataly_w [17]
Cl2(g) -------> Cl-(aq) + ClO-(aq) 

2e- + Cl2(g) -------> 2Cl-(aq) [reduction] 

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation] 
______________________________________... 
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
4 0
3 years ago
The pressure of 5.00 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant tempe
Vsevolod [243]
(P1)(V1)=(P2)(V2)
(1.50)(5.00)=(1240/760)(V2)
(7.5)/(1240/760)=V2
V2=4.596774194 L
5 0
3 years ago
Read 2 more answers
PLEASE HELP CHEMISTRY
eduard

When sodium carbonate is dissolved in water, the equation is Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq).

When carbon dioxide is placed in water, aqueous carbon dioxide is formed:  CO_2 (g) --- > CO_2(aq)

<h3>Dissolution of compounds in water</h3>

Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.

Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.

While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.

Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: CO_2 (g) --- > CO_2(aq)

More on the dissolution of substances can be found here: brainly.com/question/28580758

#SPJ1

7 0
1 year ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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