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1 year ago

6

Which event is an example of an endothermic reaction?.

1 answer:

g100num [7]1 year ago

7 0

For a reaction to be endothermic we have a process where heat is absorbed from a product and the end result is the product being colder than it originally was.

An event that can be a good practical example is using an ice pack on your arm. Your arm is warm before an ice pack is applied. After you apply an ice pack and leave it there for a while and then take it off, the site where the ice pack was would be colder than it was before the ice pack. Hope this helps!

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The chemical formula for aluminum hydroxide is Al(Oh)3. How many hydrogen atoms are present in 1 molecule here?

grandymaker [24]

There are three molecules of hydrogen.

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3 years ago

The diagram below shows different layers of sedimentary rocks.

Flura [38]

<span>Picture showing seven layers of rocks of different colors labeled A, B, C, D, E, F, and G from top to bottom;
A and B are parallel horizontal layers at the top of the diagram;
C, D, E, F, and G are slanted layers with C closest to the surface and G at the bottom.
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The inference that is most likely correct is that (<span>C) Layer F is younger than Layer D.</span>

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2 years ago

Read 2 more answers

To what Kelvin temperature must a balloon

snow_tiger [21]

Answer:

$T_2=261.46\ K$

Explanation:

It is given that,

Original temperature, $T_1=323^{\circ}C=596.15\ K$

Original volume, $V_1=2.85\ L$

We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.

According to Charles law, at constant pressure, $V\propto T$

It would means, $\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$

T₂ = ?

$T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K$

So, the new temperature is 261.46 K.

4 0

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

A sample of juice has a pH of 4.2. Whsat is the concentration in mol/dm3 of OH in the juice?<br>

Annette [7]

Answer:

6 * 10^-4M

Explanation:

use this equation

[ H + ] = 10 − pH

6 0

2 years ago