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Paha777 [63]
3 years ago
8

What is water? pls tell fast

Chemistry
2 answers:
NikAS [45]3 years ago
4 0

The mixture of Hydrogen and oxygen. It keeps all living things alive.

Zigmanuir [339]3 years ago
4 0

It is a chemical substance made of hydrogen and oxygen. Its formula is H2O. It is needed by all organisms in order to survive.

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siniylev [52]

Answer:

It sounds fine, but it may be a bit too long. It's difficult to shorten things like this, but getting more straight to the point would give it that "catchy" feel.

Explanation:

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3 years ago
Identify any solutes and solvents present in a mixture of 0.05 mol ethanol, 1 liter water, and 0.2 g hydrogen
Varvara68 [4.7K]

Answer:

Water is the solvent

Both the ethanol and the hydrogen peroxide are the solute

Explanation:

Both the hydrogen peroxide and ethanol are sisobable in water.

There are 0.05 moles of ethanol.

1 litreof water contains 55.55 moles of water.

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How does chemistry affect our world?
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Answer:

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Listed below are blood groups of o, a, b, and ab of randomly selected blood donors (based on data from the greater new york bloo
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3 0
3 years ago
Read 2 more answers
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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