Ask question

Ask question

All categories

Juliette [100K]

3 years ago

10

A doctor orders 4.5 mL of phenobarbital elixir. If the phenobarbital elixir is available as 20. mg per 7.5 mL, how many milligra

ms is given to the patient?

1 answer:

sergij07 [2.7K]3 years ago

8 0

In order to find how many milligrams is given to the patient we need to determine the rate by dividing what is available then multiply by the amount that was ordered.

Step 1: (20.0mg / 7.5mL) = 2.66667 mg/mL
Step 2: (2.66667mg/mL x 4.5mL) = 12mg
*Note that mL unit cancels out so we are left with mg

Answer: 12.0 mg of phenobarbital elixir

You might be interested in

How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate

zysi [14]

Answer:

12

Explanation:

the 4 by the element symbol O multiplied by the 3 on the outside of the parentheses

5 0

3 years ago

Read 2 more answers

Calculate the volume of liquid in the tank sketched below. Give your answer in liters, and round to the nearest 0.1 L.

wariber [46]

Answer:

1.3 L

Explanation:

The volume of a rectangular cube can be calculated using the following formula:

Volume (L) = length (cm) x width (cm) x height (cm)

Keep in mind that 1 L = 1,000 cm³.

Before you can plug the values into the equation, you need to make sure they all have the same unit. Since the length is in meters (m), you need to first convert it to centimeters.

1 meter = 100 cm

0.159 m 100 cm
--------------- x ---------------- = 15.9 cm
1 m

Now, you can solve for the volume. To find the answer is the unit liters, you need to divide the volume by 1,000.

Volume = l x w x h

Volume = 15.9 cm x 10.5 cm x 7.7 cm

Volume = 1,285.5 cm³

Volume = 1.2855 L ------> Volume = 1.3 L

7 0

2 years ago

From this list, select the element that forms π bonds most readily. from this list, select the element that forms bonds most rea

Finger [1]

Nonmetals which are located in the second row form pi bonds more easily than the elements situated in the third row and below. Actually there are no compounds or molecules known that forms covalent bonds to the noble gas Ne and Ar. Hence the other second row element which is Carbon, is the element that forms pi bonds most readily.

Answer:

<span>C</span>

3 0

3 years ago

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago