If a bar of motilium cost .0.2 dollars how much will 200 bars cost

PLEASE HURRY AND SHOW WORK

IgorC [24]

Answer:

The answer to your question is P = 1.64 atm

Explanation:

Data

Volume = 2.5 x 10⁷ L

Temperature = 22°C

Pressure = ?

Moles = 1.7 x 10⁶

R = 0.082 atm L/ mol°K

Process

1.- Convert temperature to °K

Temperature = 22 + 273

= 295°K

2.- Use the Ideal gas law to solve this problem

PV = nRT

- Solve for P

P = nRT / V

- Substitution

P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷

- Simplification

P = 41123000 / 2.5 x 10⁷

- Result

P = 1.64 atm

3 0

3 years ago

IF YOU KNOW THE REST OF THE TEST'S ANSWERS PLEASE SHARE

Semenov [28]

Answer:

B. a state in which the forward and reverse reactions are proceeding at equal rates

Explanation:

"Chemical equilibrium is the state of a chemical system at which a constant concentration of products and reagents is present. Reactions, which take place in homogeneous solutions, seem to have come to rest because no changes in concentrations of the participating substances can be determined. Substance turnover occurs only on the particle level, which is why chemical equilibrium is also referred to as dynamic equilibrium."

3 0

3 years ago

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8) Determine whether mixing each pair of the following results in a buffera. 100.0 mL of 0.10 M NH3 with 100.0 mL of 0.15 MNH4Cl

Marina86 [1]

Answer:

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

Explanation:

A buffer system is formed in 1 of 2 ways:

A weak acid and its conjugate base.

A weak base and its conjugate acid.

Determine whether mixing each pair of the following results in a buffer.

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.

b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.

NO. HCl is a strong acid and NaOH is a strong base.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).

d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.

NO. Both are bases.

6 0

3 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

Compare the molecular shape and hybrid orbitals of PF3 and PF5 molecules

MaRussiya [10]

PF3 is phosphorustrifluoride with p in the center and three f bonding with it.PF5 is phosphoruspentafluoride with p in the middle and 5 f linking to bond

5 0

3 years ago

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