M=40g V=31cm^3 what is the density

Which of these data points is the most reliable?

Sonja [21]

Answer:

third point

Explanation:

omniscient because it gives info about every character instead of one

hope this helps

5 0

2 years ago

What does ClF3 stand for

Nostrana [21]

Answer:

CHLORINE TRIFLUORIDE

Explanation:

3 0

3 years ago

If I mix bleach and namona and throw it on a person what will happen

satela [25.4K]

Answer:

They have the chance to inhale toxic fumes secreted by the mixture.

Explanation:

7 0

3 years ago

You have collected a tissue specimen that you would like to preserve by freeze drying. To ensure the integrity of the specimen,

Reika [66]

Answer:

The maximum pressure is 612.2 Pa

Explanation:

The pressure of the ice (P1) = 624 Pa

The temperature of the ice = 273.16 K

The maximum temperature the specimen = - 5 oC

= -5 + 273 = 268K

The maximum Pressure the freeze drying can be will be (P2) = ?

Using Pressure law, which shows the relationship between pressure and temperature.

P1 / T1 = P2 / T2

P2 T1 = P1 T2

P2 = P1 T2 / T1

P2 = 624 × 268 / 273.16

P2 = 612.2 Pa

The maximum pressure at which drying can be carried out is 612.2 Pa

Check the attached document more explanation. jjjjggggg

5 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago