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2 years ago

Which formula represents the compound commonly known as phosphine?

Chemistry

2 answers:

Damm [24]2 years ago

4 0

Answer:

The answer is: C) ph3.

Explanation:

Phosphine (PH3) is a colorless gas at room temperature and very toxic. It is odorless when in a pure state in concentrations of up to 282 mg / m3 (200 ppm). Phosphine is poorly soluble in water, being soluble at 2.5 - 15% v / v in most organic solvents at room temperature.

The answer is: C) ph3.

sineoko [7]2 years ago

3 0

Your answer should be c ph3

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How many grams of iron(III) oxide would be required to make 187 g of iron

enyata [817]

Answer:267 g

Explanation:

4 0

3 years ago

Atoms with loosely held valence electrons have 1. high ionization energy and high electron affinity. 2. high ionization energy a

EastWind [94]

Answer:

Explanation:

Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.

The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low

3 0

3 years ago

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago

Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det

mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0

2 years ago

How much heat must be transferred to 55 g of ice to change the ice's temperature from -13°C to -5.0°C? (The specific heat capaci

notka56 [123]

Answer:

Hey i don't know if you still need the answer but this is what i think it is:

Explanation:

B is correct answer

15 j

(Sorry if its wrong)

8 0

2 years ago

Read 2 more answers