Cathodic protection of a metal pipe against corrosion usually entails __________.

The patient needs to receive 285.71 ml.
1000 ml contains 70 gr of glucose.
   x          contains 20 gr of glucose.

x=1000*20/70

Answer: 1.14

Explanation:

To find the molarity of the acid, we will utilize the equation derived from the neutralization process:

where

are the n-factor, molarity, and volume for the acid and

represent the n-factor, molarity, and volume for NaOH.

We know that:

By substituting the known values into the equation, we get:

To determine the pH of gastric juice:

The molarity amounts to = 0.072

Thus, the pH level of the gastric juice is 1.14

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

Response:

2200 mg of antibiotic

Explanation:

The prescribed antibiotic dosage is 40 mg/kg of body weight.

For a patient weighing 55 kg, we calculate the dose of antibiotic as follows:

If we analyze 40/1000000, we can determine antibiotic allocation in kg per kg of weight

= 0.00004 kg of antibiotic for each kilogram

0.00004 multiplied by 55 (to find out the required amount for a 55 kg individual)

= 0.0022 kg

This 0.0022 figure converts to milligrams as follows

0.0022*10^6

= 2200 mg of antibiotic is indicated for a patient weighing 55 kg.

Given: Mass of the ionic compound = 10.00 g Mass of water = 75.0 g Initial temperature of water T1= 23.2 C Final temperature of water T2 = 31.8 C Specific heat of water c = 4.18 J/gC To determine: Enthalpy of dissolution of the ionic compound Heat gained by water equation: Q = mcΔT m = mass of water c = specific heat ΔT = change in temperature (T2-T1) Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution) Therefore, q(ionic) = 2696 J ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g Answer: A) enthalpy change = 2.7*10² J/g