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Volgvan
3 years ago
14

How many molecules are in 1.00 g of C10H14N2? 3.71 x 1021 9.76 x 1025 1.02 x 10-26 2.32 x 1022

Chemistry
1 answer:
Reika [66]3 years ago
8 0

Answer:

3.71 x 1021 (First option)

Explanation:

Let's determine the molar mass of the compound.

C₁₀H₁₄N₂ = 12. 10 + 14 .1 + 14 .2 = 162 g/mol

So 162 grams of compound are contained in 1 mol

1 g of  C₁₀H₁₄N₂ are contained in ( 1 / 162) = 0.00617 moles

Let's conver the moles into amount of molecules (mol . NA)

0.00617 . 6.02×10²³ = 3.71×10²¹ molecules

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Answer:

well the world would be shocked

Explanation:

and suprise

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3 years ago
When filtered through a funnel into a flask, a mixture of substances X, Y and Z gets separated as below: - X stays in the funnel
alisha [4.7K]

Z is the solvent, Y is soluble in water while X is insoluble in water.

<h3>Filtration</h3>

Filtration is a method of separation of substances based on particle size. Only a particular particle size can pass through the filter. The substance that remains in the filter is the residue while the substances that passes through the filter is called the filtrate.

From the observation in the question Z is the solvent, Y is soluble in water while X is insoluble in water.

Learn more about separation of mixtures: brainly.com/question/863988

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3 years ago
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4 0
3 years ago
Calculate the molarity of a sugar solution if 4 liters of the solution contains 12 moles of sugar.
castortr0y [4]

Answer:

3M

Explanation:

Molarity is one of the measures of the molar concentration of a solution, which can be calculated by using the formula below:

Molarity = number of moles ÷ volume

From the information given in this question, 4 liters of a solution contains 12 moles of sugar. This means that n = 12mol and V = 4L

Molarity = n/V

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8 0
3 years ago
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
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