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svp [43]
3 years ago
13

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

Chemistry
1 answer:
boyakko [2]3 years ago
7 0

Answer:

lo siento, no sé punto libre. :p

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What is the difference between mass and weight in detail?
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Answer:

Explanation:

Hope it helps

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5 0
4 years ago
if your total amout of liquid in milliliters is 10 and you total amout of water in milliliters is 7.what is the consentration of
Dominik [7]

3 concentration of the water

7 0
3 years ago
Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

7 0
3 years ago
You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid
Art [367]

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

Best regards.

5 0
3 years ago
Which of the following is a scientific question you could ask about the game
aniked [119]

Answer:

Answer is option C

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