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tino4ka555 [31]
3 years ago
8

Which of these is an equation that follows the law of conservation of mass

Chemistry
1 answer:
Vilka [71]3 years ago
5 0

Answer:

D

Explanation:

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Answer:

You may, but it is too risky.

Even though you are being cautious around using electric equipment around water, you'll never know what can happen. You might accidentally drop that piece of electrical equipment you are using into the water. Water can be splashed around by someone or something without you noticing it and it may affect the object you are using. Sometimes, if water comes in contact with an electrical object, it may cause you electric shocks or the equipment you are using has a chance of exploding and may hurt you. You can guarantee that waterproof electrical equipment is safe to use, but it is better not to risk it too much.

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3 years ago
Which element would you expect to behave the most like carbon (c)? apex?
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3 years ago
Read 2 more answers
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
2 years ago
Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r
7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

7 0
3 years ago
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