All categories

3 years ago

_____ sedimentary rocks are made of once-living things or their pieces. A. Organic B. Inorganic C. Chemical D. Fragmental

Chemistry

2 answers:

kramer3 years ago

8 0

I believe the answer is A. Hope this helps.

lawyer [7]3 years ago

7 0

Answer:

A. Organic

Explanation:

Once living things have rocks that are made up of organic content as sedimentary rocks are best suited to preserve the remains of the fossils and for keeping records.

The organic sediments in rocks are formed due to the accumulation of organic debris collected from various natural sources. As there are swampy sediments that contain various carbon traces and are black, and soft.

Fossils are very important elements of depositional environment and especially trace fossils like burrows, footprints and leaf impressions, etc. Some examples of corals and other shell fossils indicate marine deposition.

You might be interested in

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

3 years ago

The rays produced in a cathode tube are

wel

Answer:

Electrons

Explanation:

Cathode rays carry electronic currents through the tube. Electrons were first discovered as the constituents of cathode rays. J.J. Thomson used the cathode ray tube to determine that atoms had small negatively charged particles inside of them, which he called “electrons.”

7 0

3 years ago

Read 2 more answers

Someone help me pls will make you as brain

jonny [76]

Answer:

i, ii, and iii

Explanation:

3 0

3 years ago

Read 2 more answers

Would an opah fish be better suited to live in cold water or warm water? explain why

Kaylis [27]

Don't blame me if you get it wrong i'm a dumb but yet successful student

Warm blood gives deep-sea fish a boost, according to Wegner. The opah's muscles and nervous system likely function faster than an equivalent fish with cold blood. ... This fish, the southern opah, lives in colder waters than the northern opah, so it would be harder to keep warm, Wegner said — but even more beneficial.

4 0

3 years ago

Please help me with this chemistry problem

Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0

3 years ago