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Dimas [21]
3 years ago
9

In a certain medical test designed to measure carbohydrate tolerance, an adult drinks 6 ounces of a 40% glucose solution. When t

he test is administered to a child, the glucose concentration must be decreased to 30%. How much 40% glucose solution and how much water should be used to prepare 6 ounces of 30% glucose solution?
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
8 0

Answer:

4.5 ounces of 40% glucose solution and 1.5 ounces of water.

Explanation:

The new solution will be prepared by adding more solvent, so the relation between concentration and volume is given by:

C1xV1 = C2xV2

Where C is the concentration, V is the volume, 1 represents the initial solution and 2 the final solution.

We want 6 ounces of a solution which 0.3 glucose, so :

0.4xV1 = 0.3x6

0.4xV1 = 1.8

V1 = 4.5 ounces

So, it's necessary 4.5 ounces of the 40% glucose solution. The volume of water (Vw) will be the total volume less the volume of the solution:

Vw = 6 - 4.5

Vw = 1.5 ounces

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The combination of potassium-sparing diuretics and salt substitutes can result in dangerously high blood levels of:
alisha [4.7K]

Answer:

b. potassium.  

Explanation:

Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.

Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.

4 0
3 years ago
If the amount of Adams if each type on the left and right sides of a reaction difference what must be done to balance it
Ivanshal [37]

Hey there,

Your answer would be

Coefficients are placed in front of the reactants and/or products

Hope this helps,

<h2>- <em>Mr. Helpful</em></h2>

8 0
3 years ago
A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill
nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

5 0
3 years ago
A certain half-reaction has a standard reduction potential -1.33V . An engineer proposes using this half-reaction at the anode o
Paha777 [63]

Answer:

a. -0.63 V

b. No

Explanation:

Step 1: Given data

  • Standard reduction potential of the anode (E°red): -1.33 V
  • Minimum standard cell potential (E°cell): 0.70 V

Step 2: Calculate the required standard reduction potential of the cathode

The galvanic cell must provide at least 0.70V of electrical power, that is:

E°cell > 0.70 V    [1]

We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.

E°cell = E°cat - E°an   [2]

If we combine [1] and [2], we get,

E°cat - E°an > 0.70 V

E°cat > 0.70 V + E°an

E°cat > 0.70 V + (-1.33 V)

E°cat > -0.63 V

The minimum E°cat is -0.63 V and there is no maximum E°cat.

3 0
3 years ago
The reacted side of a balanced chemical equation is shown below. C3H8 + 5O2 How many oxygen atoms should there be on the product
atroni [7]
10 atoms. If there are 10 in the reactants you need the same number in the products
5 0
3 years ago
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