How many milliliters of a 3.0 M HCL solution are required to make 250.0 milliliters of 1.2 M HCL

1. A cube of iron at 20.°C is placed in contact with a cube of copper at 60.°C. Which statement describes the initial flow of he

4vir4ik [10]

Answer:

D) Heat flows from the copper cube to the iron cube.

Step-by-step explanation:

The reason comes from the second law of thermodynamics:

Heat energy will flow spontaneously from an area of high temperature to one of low temperature.

When you bring two objects of different temperature into contact, heat energy always moves from the hotter to the cooler object.

The copper cube is hotter than the iron, so heat flows from the copper cube to the iron cube.

here hope this helps

4 0

3 years ago

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How many liters of gaseous hydrogen bromide at 58°C and 0.891 atm will a chemist need if she wishes to prepare 3.50 L of 1.20 M

Taya2010 [7]

Answer:

15.4 l

Explanation:

5 0

3 years ago

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78.2 divided by 32cm³

klio [65]

Answer:

2.44375

rounded=2.44

6 0

3 years ago

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Tracie measured 87.47 mg of cholesterol in 0.03 mL of blood. What is the density of this mixture in g/mL?

Lelu [443]

We can use a simple equation to solve this problem.

d = m/v

Where d is the density, m is the mass and v is the volume.

d = ?

m = 87.47 mg = 87.47 x 10⁻³ g

v = 0.03 mL

By applying the equation,

d = 87.47 x 10⁻³ g / 0.03 mL

d = 2.92 g/mL

Hence, the density of the mixture is 2.92 g/mL.

5 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago