All categories

3 years ago

why do we heat the test tube in a glass beaker during a chemistry experiment and not directly over a flame?

1 answer:

7 0

It is not directly over a flame because it depends on the substance you might not want to heat it too much.you never know what could happen

You might be interested in

ASAP PLLEEASE HURYYYYYYYYYYRRRRRRRRRRRRYY (NO FILES FOR ME NEEDING TO DOWNLAOD PLEASE!)

pickupchik [31]

Answer:

1.Fill in the blank with the correct term to make the statement true.

During the (night)

at the beach, the land cools down

(slower)

than the

(ocean).

Because of this, the air over the land is

(Warmer) than the air over the ocean. The

(More)

dense air rises over the ocean and the cooler,

(More)

dense air blows in from over the land creating a

(sea)

breeze.

2.Which statement is true about winds?

Winds always blow from low to high pressure.

true

Winds always blow from water to land.

false

Winds always blow from land to water.

false

3. Winds always blow from high to low pressure. How do landmasses impact the movement of air and water?

They stop all movement of both air and water.

They speed up the air and water and allow it to pass though.

They change the direction and slow them down.

think this is it

They slow down the air but speed up the water.

pt:2

answering the pictures

9- I don know sorry

15-B

16-A

Explanation:

just trying to help LET ME KNOW IF IT'S WORNG

3 0

2 years ago

Read 2 more answers

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit

kati45 [8]

<u>Answer:</u> The $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

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

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}$

We are given:

$p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm$

Putting values in above equation, we get:

$K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = $5.34\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol$

Hence, the $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

5 0

2 years ago

What are the two main ways that minerals are classified

Kazeer [188]

country of origin and chemical makeup

3 0

2 years ago

A decomposition reaction, with a rate that is observed to slow down as the reaction proceeds, has a half-life that does not depe

mr_godi [17]

Answer: A plot of the natural log of the concentration of the reactant as a function of time is linear.

Explanation:

Since it was explicitly stated in the question that the half life is independent of the initial concentration of the reactant then the third option must necessarily be false. Also, the plot of the natural logarithm of the concentration of reactant against time for a first order reaction is linear. In a first order reaction, the half life is independent of the initial concentration of the reactant. Hence the answer.

3 0

3 years ago

You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

bonufazy [111]

Answer:

$\frac{[F^{-}]}{[HF]}$ is larger