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3 years ago

A bottle contains 175 ml of insulin at a concentration of 40.0 mg/ml . what is the total mass of insulin in the bottle?

Chemistry

1 answer:

Inessa [10]3 years ago

5 0

D - density: 40 mg/ml
V - volume: 175 ml
m - mass: ??
__________
d = m/V
m = V×d
m = 175×40
m = 7000 mg = 7 g

:•)

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

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

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

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Sladkaya [172]

Money so i can get a nice body with the money hahah

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Which of the following would not be an isotope of carbon-12? A. Carbon-14 B. Carbon-13 C. Boron-12 D. Carbon-11

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What is the enthalpy of formation (?H?f) of NaHCO3(s) from its constituent elements.

vova2212 [387]

Answer:

Enthalpy of formation = -947.68KJ/mol

Explanation:

Enthalpy of formation is the heat change when one mole of a substance is formed from its element in its standard states and in standard conditions of temperature and pressure. it may be positive or negative, if positive, it is an endothermic reaction where the heat content of the product is greater than that of the reactants, and if negative, it is exothermic reaction - where the heat content of the reactants is greater than the products. the enthalpy of formation is measured in KiloJoule/Moles (KJ/Mole).

From the value of the enthalpy of formation of NaHCO3, it shows that the reaction is exothermic, that is the formation of NaHCO3 from its constituents elements. As such, the heat content of the reactants is greater than the products.

The step by step explanation is shown in the attachment.

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