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2 years ago

How do you know something is there if you can't see it?

Chemistry

2 answers:

gavmur [86]2 years ago

5 0

Answer:

we know that someone or something is near us even when we cant see it you see our self conscious minds can pick up on the slightest detail even when we dont pick up on it also you can thank the six senses as well for why we can tell something or someone is there

Explanation: i watch criminal minds and pay attention to the random facts spencer reid spits out

aliina [53]2 years ago

3 0

Answer:

Explanation:

Use your other senses such as touch or smell, does it react in different temperatures, is it then visible under a microscope, does it react with different elements or gases? Think outside the box.

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Which substance can be broken down by a chemical change

ELEN [110]

Lots of substances. Rocks, metals, minerals, you name it.

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2 years ago

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Write any three acids which are used in our daily life. Also, write an application of each

lidiya [134]

Sodium fluoride- to brush teeth
Citric acid- orange juice for breakfast
Sodium hydroxide- cleaning agent

3 0

2 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of stomach acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

Hence, the volume of HCl neutralized is 1.25 mL

3 0

3 years ago

compare the boiling points of propane (C3H8) with that of (C4H10) .. explain your answer ...plzzz guys help me with this

yawa3891 [41]

i may be wrong but Use the normal boiling points: propane, C3H8, –42.1˚C; butane, C4H10, –0.5˚C; pentane, C5H12, 36.1˚C; hexane, C6H14, 68.7˚C; heptane, C7H16, 98.4˚C; to estimate the normal boiling point of octane, C8H18.

7 0

3 years ago

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