Ask question

Ask question

All categories

3 years ago

13

Identify fossil types match each each type of fossil to its description

2 answers:

GarryVolchara [31]3 years ago

5 0

Answer: you welcome

Explanation:

dem82 [27]3 years ago

4 0

Well we need to see the fossil

You might be interested in

The entropy of an exothermic reaction decreases. This reaction will be spontaneous under which of the following temperatures?

kakasveta [241]

Answer:Low temperatures

Explanation:

∆G= ∆H-T∆S

If ∆H is negative (exothermic reaction), then in order to maintain ∆G<0 which is the condition for spontaneity; T must decrease. This is because, decrease in T will keep the difference of ∆H and T∆S at a negative value in order to satisfy the above stated condition for spontaneity.

3 0

4 years ago

Read 2 more answers

Nucleic acids are chains of 5-carbon sugars linked by ____ bonds with an organic base protruding from each sugar.

Mekhanik [1.2K]

<span>B) phosphodiester </span> is the correct answer

3 0

4 years ago

How much Oxygen would be required to react with 101 g of Methane?

AlexFokin [52]

Answer:

404g

Explanation:

6 0

3 years ago

For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of

torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For $Al_2O_3$

Mass of $Al_2O_3$ = 4.07 g

Molar mass of $Al_2O_3$ = 101.96 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.07\ g}{101.96\ g/mol}$

$Moles\ of\ Al_2O_3= 0.0399\ mol$

According to the reaction:

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

1 mole of $Al_2O_3$ on reaction produces 1 mole of $Al_2(SO_4)_3$

So,

0.0399 mole of $Al_2O_3$ on reaction produces 0.0399 mole of $Al_2(SO_4)_3$

Moles of $Al_2(SO_4)_3$ obtained = 0.0399 mole

Molar mass of $Al_2(SO_4)_3$ = 342.2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0399= \frac{Mass}{342.2\ g/mol}$

$Mass= 13.7\ g$

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

$\%\ yield =\frac{10.4}{13.7}\times 100$

<u>% yield =76 %</u>

6 0

3 years ago

You lab measurements are 23.29, 23.1g, and 23.2g for each trial. Which of the following are your measurements?

STatiana [176]

Answer:

Precise and maybe accurate.

Explanation:

To know if it is accurate, determine whether or not your measurements are similar to the correct measurements

3 0

3 years ago