All categories

3 years ago

What reaction would you expect chromium to have with water

1 answer:

8 0

Answer: It does not react with water, but reacts with most acids. It combines with oxygen at room temperature to form chromium oxide. Chromium oxide forms a thin layer on the surface of the metal, protecting it from further corrosion (rusting).

Explanation:

You might be interested in

A gas has v= 4.0L, T =27 degree C and P=2.0 atm. What is the V if T=327 degree C and P=8.0 atm?

wariber [46]

As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D

3 0

4 years ago

For the following reaction, 2.45 grams of methane (CH4) are allowed to react with 27.7 grams of carbon tetrachloride . methane (

rosijanka [135]

Answer:

The limiting reactant is CH₄

26.0g of CH₂Cl₂ is the maximum amount that can be formed

4.15g CCl₄ will remain

Explanation:

The reaction of methane, CH₄, with carbon tetrachloride, CCl₄ is:

CH₄ + CCl₄ → 2CH₂Cl₂

To find the maximum mass of dichloromethane that can be determined we need to find moles of methane and carbon tetrachloride:

Moles CH₄:

2.45g * (1mol / 16.04g) = 0.153 moles

Moles CCl₄:

27.7g * (1mol / 153.82g) = 0.180 moles

That means just 0.153 moles of CCl₄ will react until CH₄ is over.

The limiting reactant is CH₄

Assuming the whole 0.153 moles will react, the moles of CH₂Cl₂ will be:

0.153 moles CH₄ * (2 moles CH₂Cl₂ / 1 mole CH₄) = 0.306 moles of CH₂Cl₂

The mass is (Molar mass dichloromethane: 84.93g/mol):

0.306 moles of CH₂Cl₂ * (84.93g / mol) = 26.0g of CH₂Cl₂

The moles of CCl₄ that remain are:

0.180 moles - 0.153 moles = 0.027 moles

In grams:

0.027 moles * (153.82g / mol) = 4.15g CCl₄

3 0

3 years ago

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol

diamong [38]

Answer: ksp= 4s³

Explanation:

4 0

3 years ago

You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam

katovenus [111]

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression $d=\frac{m}{V}$ where m is the mass and V is the volume, therefore:

- Density of the cube A:

$d_{A}=\frac{m_{A}}{V_{A}}$

- Density of the cube B:

$d_{B}=\frac{m_{B}}{V_{B}}$

- Density of the cube C:

$d_{C}=\frac{m_{C}}{V_{C}}$

Solving for mass:

$m_{A}=d_{A}*V_{A}$

$m_{B}=d_{B}*V_{B}$

$m_{C}=d_{C}*V_{C}$

And all the three cubes have the same mass, so:

$m_{A}=m_{B}=m_{C}$

Therefore:

$d_{A}*V_{A}=d_{B}*V_{B}$ (Eq.1)

$d_{A}*V_{A}=d_{C}*V_{C}$ (Eq.2)

Solving for $d_{1}$ in Eq.1:

$d_{A}=d_{B}\frac{V_{B}}{V_{A}}$

Replacing values for the volume:

$d_{A}=d_{B}\frac{16.7mL}{25.9mL}$

$d_{A}=d_{B}*0.64$

As we know the density of the aluminum is $2.7\frac{g}{cm^{3}}$ , so replacing this value for $d_{B}$ :

$d_{A}=2.7\frac{g}{mL}*0.64$

$d_{A}=1.728\frac{g}{mL}$

that is the density of the magnesium.

Solving for $d_{C}$ in Eq.2:

$d_{C}=d_{A}\frac{V_{A}}{V_{C}}$

$d_{C}=d_{A}\frac{25.9mL}{4.29mL}$

$d_{C}=d_{A}*6.04$

$d_{C}=1.728\frac{g}{mL}*6.04$

$d_{C}=10.4\frac{g}{mL}$

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

8 0

4 years ago

How many ppm is 2.372 ppb?

rewona [7]

Answer: 1.314 ppm
explain: thats what someone said

4 0

3 years ago