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sergeinik [125]
3 years ago
10

I don’t know how to solve this

Chemistry
1 answer:
stepladder [879]3 years ago
7 0
Hi,

Answer is 191.2.

800J = 191.2 cal

Hope this helps.
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Cho biết độ tan của NH4Cl trong nước ở 20oC và 70oC lần lượt là 37,2 g/100 gam nước và 60,2 gam/100 g nước. Hòa tan 166,8 gam NH
Stolb23 [73]

Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C

Actini(III) hydroxide Ac(OH)3   0,0022    

Amonia NH3 1176 900 702 565 428 333 188

Amoni azua NH4N3 16  25,3  37,1  

View 42 more rows  

                    hehe

8 0
2 years ago
Why does mecury hardly evaporate in room temperature
Crazy boy [7]
Because elemental mercury is a liquid at a room temperature
6 0
2 years ago
A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
I need helpppppppp, please
NNADVOKAT [17]

Explanation:

cant answer without context

8 0
2 years ago
This is about rocks and/or minerals
vampirchik [111]
I need the same answer
5 0
2 years ago
Read 2 more answers
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