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MissTica
3 years ago
12

If you wanted to prepare a 15.58 L of a 6.25 M solution, what volume of a 14.73 M solution must you start with?

Chemistry
1 answer:
Fed [463]3 years ago
4 0

Answer:

6.61 L

Explanation:

M1 = 6.25

V1 = 15.58

M2 = 14.73

M1V1 = M2V2

(6.25)(15.58) = (14.73)V2

V2 = 6.61 L

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answer is potasium means k

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What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
Consider the reaction cacn2 + 3 h2o → caco3 + 2 nh3 . how much caco3 is produced if 47.5 moles nh3 are produced? 1. 0.950 g 2. 9
zhannawk [14.2K]
The  CaCO3  produced  if  47.5   moles of  NH3  produced  is   calculated  as  follows

CaCN2  +3H2O  = CaCO3  + 2NH3

by  use   of  mole  ratio  between  CaCO3 to NH3  which  is 1:2  the  moles  of  CaCO3  is therefore =  47.5 /2= 23.75  moles

mass of CaCO3   is  therefore = moles  x  molar  mass
= 23.75 moles  x  100g/mol=  2375 grams   which  is approximate   2380  grams(answer 6)
8 0
4 years ago
What is/are the product(s) of a neutralization reaction of a carboxylic acid? View Available Hint(s) What is/are the product(s)
lutik1710 [3]

Answer:

A carboxylate salt and water

Explanation:

A carboxylic acid is an organic compound that has general formula RCOOH, where R is a carbon chain. Because it's an acid, the neutralization will happen when it reacts with a base, such as NaOH.

When this reaction occurs, the base will dissociate in Na⁺ and OH⁻, and the acid will ionize in RCOO⁻ and H⁺, so the products will be RCOO⁻Na⁺ (a carboxylate salt) and H₂O (water).

7 0
3 years ago
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