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2 years ago

How much time is needed to form most fossils? Help please!

Chemistry

2 answers:

Varvara68 [4.7K]2 years ago

5 0

Answer:millions of years

Explanation:

Edge 2021

Hunter-Best [27]2 years ago

3 0

Answer:

a minimum of 10,000 years

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Calculate the total number of days therapy available within a 300 ml bottle of ranitidine (as hydrochloride) 75 mg/5 ml. oral so

Lisa [10]

Answer: This therapy is available for 15 days.

Explanation:

We are given:

Oral solution dosage = 75 mg/5 mL

To calculate the volume of oral situation for single dose per, we use unitary method:

The volume required for 75 mg of solution is 5 mL

So, the volume required for 300 mg of solution will be = $\frac{5mL}{75mg}\times 300mg=20mL$

The total volume of the ranitidine bottle = 300 mL

To calculate the number of days, we divide the total volume of the bottle by the volume of dose taken per night, we get:

$\text{Number of days}=\frac{\text{Total volume}}{\text{Volume of dose taken per night}}=\frac{300mL}{20mL}=15$

Hence, this therapy is available for 15 days.

8 0

3 years ago

Find the wavelength of the electromagnetic radiation with a frequency of 3x10^-18Hz (Show Work)

aalyn [17]

Answer:

λ = 1×10²⁶m

Explanation:

Given data:

Wavelength of radiation = ?

Frequency of radiation = 3×10⁻¹⁸Hz

Solution:

Formula:

c = f × λ

c = speed of wave = 3×10⁸ m/s

by putting values,

3×10⁸ m/s = 3×10⁻¹⁸Hz × λ

λ = 3×10⁸ m/s / 3×10⁻¹⁸s⁻¹

λ = 1×10²⁶m

5 0

3 years ago

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne

mars1129 [50]

163 grams (3 sig. fig.).

<h3>Explanation</h3>

Formula of carbon(IV) oxide (a.k.a. carbon dioxide): $\text{CO}_2$ .
Molar mass of $\text{CO}_2$ : $\underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}$ .

Formula of ethyne: structural $\text{H}-\text{C}\equiv\text{C}-\text{H}$ or molecular $\text{C}_2\text{H}_2$ .
Molar mass of $\text{C}_2\text{H}_2$ : $2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}$ .

All carbon atoms in that 104 grams of ethyne will end up in $\text{CO}_2$ . Number of moles of molecules in 104 grams of ethyne:

$n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}$ .

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

$n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}$ .

There are one carbon atom in each $\text{CO}_2$ molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make $n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol}$ of $\text{CO}_2$ .