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Serjik [45]
2 years ago
7

How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

Chemistry
1 answer:
balandron [24]2 years ago
7 0

Answer:

0.375mL of solution of nalorphine must be injected

Explanation:

A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:

1.5x10⁻³g * (100mL / 0.40g) =

<h3>0.375mL of solution of nalorphine must be injected</h3>
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How many grams of hydrogen reacts with 32 g of oxygen to produce 34 g of hydrogen peroxide
nordsb [41]

Answer:

2 grams.

Explanation:

H2 + O2 --->  H2O2

Using molar masses:

2*1 g hydrogen   reacts with 2*16  g oxygen.

so 2g H2 reacts with 32 g O2.

8 0
3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
2 years ago
3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measur
lilavasa [31]

Answer:

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) +  8O₂(g)  → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 =  1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

3 0
3 years ago
Read 2 more answers
Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of
Oxana [17]
<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Al→3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>
7 0
3 years ago
3.Which of these is NOT part of the cell theory? * "
Ymorist [56]

Answer:

Cells are the smallest living unit of an organism

8 0
3 years ago
Read 2 more answers
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