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Juli2301 [7.4K]
3 years ago
8

As a purchasing agent for a pharmaceutical company, how much chlorine, Cl2, do you need to order to react completely with 500 kg

of platinum, Pt, to make cisplatin, PtCl2(NH3)2?
Chemistry
1 answer:
Aleksandr [31]3 years ago
3 0
The formula tells you the proportion in mols

2 mols of Cl: 1 mol of Pt

Now use the atomic masses:

Cl: 35.5 g/mol
Pt: 195 g/mol

Then you have these equation:

[2*35.5 g Cl] / 195 g Pt = x / 500 kg Pt

Solve for x.

x = (71 g Cl / 195 g Pt) * 500 kg Pt

x = 182 Kg Cl = 182 kg Cl2.
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Which five statements about hemoglobin and myoglobin structure are true? Each iron atom can form six coordination bonds. One of
Ulleksa [173]

Answer:

Here's what I get  

Explanation:

Each iron atom can form six coordination bonds. One of these bonds is formed between iron and oxygen.  

TRUE

By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron atom.  

TRUE

Molecular oxygen binds irreversibly to Fe²⁺ in heme.  

False. It binds reversibly with heme.

Heme is composed of an organic protoporphyrin component and a metal atom.  

TRUE

Hemoglobin and myoglobin are heterotetramers.

False. Hemoglobin is a tetramer. Myoglobin a monomeric protein.

Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron atom.

TRUE

Each hemoglobin molecule can bind four oxygen molecules; each myoglobin can bind only one oxygen molecule.

TRUE

7 0
3 years ago
Read 2 more answers
Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H
solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol

2) Moles of ammonia gas=\frac{2.53 g}{17 g/mol}=0.1488 mol

Moles of oxygen gas =\frac{3.53 g}{32 g/mol}=0.1101 mol

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

\frac{4}{3}\times 0.1101 mol=0.1468 mol of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

\frac{2}{3}\times 0.1101 mol=0.0734 mol of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100

Percentage yield of the reaction:

\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%

36.37% is the percent yield of the reaction.

3 0
2 years ago
Question 4 (1 point)<br> If you have exactly one million atoms how many moles do you have?
rjkz [21]

Answer:

                       1.66 × 10⁻¹⁸  Moles

Explanation:

                     As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

             Number of Moles  =  Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

              Number of Moles  =  1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

              Number of Moles  =  1.66 × 10⁻¹⁸  Moles

7 0
3 years ago
Heat transfer between two substances is affected by specific heat and the chemical composition of the substances. state of matte
lesya [120]

Answer:

the answer is D

Explanation:

7 0
2 years ago
Read 2 more answers
Consider the half reactions below for a chemical reaction.
ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0
2 years ago
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