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3 years ago

6

Use "Folded Mountain" in a sentence? For science

1 answer:

densk [106]3 years ago

8 0

The folded mountain seemed to grow bigger and taller

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7. How many grams are there in 2.5 moles of silver?

exis [7]

2.5 moles of silver is 269.6705 grams

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3 years ago

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GUYS I NEED THIS ASAP

liberstina [14]

It’s DEFINITELY 2 like DEFINITELY

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3 years ago

A neutral atom of which of the four element has the smallest radius

arsen [322]

Answer:
Element 1

Explanation:
The ionization energy is defined as the energy required to remove electrons from the atoms.
We know that the nucleus of the atom attracts the electrons, thus, bound these electrons to the atom.
This means that as the radius decreases, the force of attraction between the nucleus and the electron will increase, therefore, the energy required to remove the electron would increase (and vice-versa).

Based on the above, the atom with the smallest radius would be the atom with the largest first ionization energy.

Hope this help :)

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3 years ago

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How do many types of cells, like bacteria, reproduce?

umka2103 [35]

Answer:

cells like bacteria are bisexual so they split and their offspring is 100 percent like the parent and this process happens over and over

Explanation:

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2 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0

3 years ago