All categories

3 years ago

Los termómetros de mercurio no pueden medir temperaturas menores a -30oC debido a que a esa temperatura el Hg

Chemistry

1 answer:

Ksivusya [100]3 years ago

3 0

Answer:

243.15 K y -22 F

Explanation:

Para convertir la temperatura desde grados Celsius (°C) a kelvin (K) se debe sumar 273.15 K:

K = °C + 273.15 = -30°C + 273.15 K = 243.15 K

Para convertir desde grados Celsius (°C) a Fahrenheit (F), se multiplica por 9/5 y se suma 32, como sigue:

F = (9/5 x °C) + 32 = (9/5 x (-30°C)) + 32= -22 F

Por lo tanto, una temperatura de -30°C corresponde a 243.15 Kelvin y -22 grados Fahrenheit.

You might be interested in

A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib

rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

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

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

7 0

4 years ago

How does skin protect humans

iVinArrow [24]

Answer:

it protects everything inside our body

Explanation:

7 0

4 years ago

Read 2 more answers

Se disuelven 10 g de acido carbónico H2CO3 en 150 g de agua. Determina la concentración % m/m de esa disolución?

sergey [27]

The question is: 10 g of carbonic acid H2CO3 are dissolved in 150 g of water. Determine the% m / m concentration of that solution?

Answer: The% m / m concentration of that solution is 6.66%.

Explanation:

Given: Mass of solute = 10 g

Mass of solvent = 150 g

Formula used to calculate the %m/m is as follows.

$Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{10 g}{150 g} \times 100\\= 6.66%$

Thus, we can conclude that the% m / m concentration of that solution is 6.66%.

6 0

3 years ago

The thikness of an eye lash is 15 mm how much this will be in meters ? write it in scientific Notation.

deff fn [24]

Answer

1.5x10^-3 m

24.91x10^5

Explanation:

15 millimeters

- 1 mm = .0015

exponent indicates which way the decimal moved and to how many places

we moved the decimal backwards (negative) 3 spaces

2491 km= 2491000 m

exponent indicates which way the decimal moved and to how many places

we moved the decimal forwards (positive) 5 spaces

2491000.00 --> 24.91000

6 0

4 years ago

A 45.0 mL solution of 0.0450 M hydroxylamine is extracted with 125 mL of solvent. The distribution constant for the reaction is

julia-pushkina [17]

Answer:

pH = 4.5, concentration = 0.045 M.

pH = 6.5, concentration = 0.175 M.

Explanation:

The ka for the can be calculated by using the formula below;

Ka = 10^-pka = 10^-5.960 = 1.1 × 10^-6

The concentration of hydrogen ion at pH = 4.50 can be calculated as given below;

{H^+ } = 10^-4.50 = 3.2 × 10^-5 M.

(NB=> 10 in this regards means the inverse of log).

The next step is to determine the distribution coefficient which can be calculated by using the formula below;

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient =( 5 × 1.1 × 10^-6 ) / 1.1 × 10^-6 + 3.2 × 10^-5 M. = 5.5 × 10^-6/ 3.2 = 0.00000171875

The fraction remaining from the compound = 45.0 mL / 45.0 mL + (0.00000171875 × 125).

= 0.999995.

Thus, the concentration at pH = 4.5 = 0.999995 × 0.0450 M = 0.045 M

(B). pH=6.50, thus the concentration of Hydrogen ion = 10^-6.5 = 3.2 × 10^-7 M.

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient = (5 × 1.1 × 10^-6)/ 1.1 × 10^-6 + 3.2 × 10^-7 M).

distribution coefficient = 5.5 × 10^-6/ 1.42 × 10^-6 = 3.9.

Therefore, the concentration = 3.9 × 0.0450 M = 0.175 M.

8 0

3 years ago