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3 years ago

consider the letters in the word chemistry. use them to make as many word " compounds " as is possible with 9 elements.

Chemistry

2 answers:

Alborosie3 years ago

7 0

Explanation:

A compound is defined as the substance which contains different type of elements that are chemically combined together in a fixed ratio by mass.

Therefore, considering the letters in the word chemistry, name of different compounds will be as follows.

From C - Copper chloride.

From H - Hydrogen peroxide

From E - Europium(III) oxide

From M - Magnesium sulfate

From I - Iodine bromide

From S - Sulfur dioxide

From T - Titanium(IV) oxide

From R - Rhodium(III) chloride

From Y - Ytterbium(III) fluoride

gladu [14]3 years ago

5 0

Cobalt
Hydrogen
Europium
Magnesium
Iodine
Sulfer
Titanium
Radon
Yttrium

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NiO is to be reduced to nickel metal in an industrial process by use of the reaction:NiO(s) + CO(g) <-----> Ni(s) + CO2(g)

slavikrds [6]

Answer:

The reaction will occur, in the forward direction, and NiO will be reduced to Ni

Explanation:

Step 1: Data given

Kp = 6.0 * 10²

Pressure of CO = 150 torr

Total pressure never exceeds 760 torr

Step 2: The balanced equation:

NiO(s) + CO(g) ⇆ Ni(s) + CO2(g)

Step 3: Calculate P(CO) and P(CO2)

For this equilibrium, the expression for Kp is:

Kp = P(CO2)/P(CO) = 6.0*10^2

(NiO and Ni are solids so do not change anything to the Kp

⇒ with P(CO) = 150 torr

⇒ P(CO) = 760 - 150 = 610 torr

Step 4: Calculate the reaction quotient

Q = 610/150 = 4.1

Q<<< Kp

There are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

This means the reaction will proceed in the forward direction, and NiO will be reduced to Ni.

6 0

3 years ago

The gram molecular mass of co2 is the same as the gram molecular mass of ?

kari74 [83]

The answer is 1. CO. A gram-molecular mass is defined as mass in grams numerically equal to the molecular weight of a substance or the sum of all the atomic masses in its molecular formula. Since CO2 and CO has both carbon and oxygen, the gram-molecular mass does not change. For a compound with carbon and oxygen, the molecular mass comes respectively from 12 (atomic mass of carbon) + (2 × 16) (atomic mass of oxygen), which is 44 g.

3 0

3 years ago

Read 2 more answers

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

Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe

Viktor [21]

There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.

7 0

3 years ago

Which of the following is an electrolyte? a. PbCl2 b. MgS c. HNO3 d. Al2(Cr2O7)3

Phoenix [80]

Answer: A: PbCl2

Hope dis helps!

8 0

3 years ago