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scoray [572]
3 years ago
10

Hello it's my new id myself numu ​

Chemistry
2 answers:
zavuch27 [327]3 years ago
7 0

Answer:

Hi numu !

How are you .

:D

Hope you have nice time

zaharov [31]3 years ago
3 0

Answer:

Hi numu I'm Tanvi

Explanation:

how r u?

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What is the coefficient in this chemical formula? 3C6H12O6
erma4kov [3.2K]

Answer:

C-3

Explanation:

7 0
3 years ago
Read 2 more answers
Explain how the removal of thermal energy (heat) from matter affects its particles
den301095 [7]

Answer:

You can change an object's state of matter by adding or removing thermal energy. When you add thermal energy to an object, these things can happen: Particles move faster

Explanation:

6 0
3 years ago
Which weather condition is NOT associated with warm fronts??
slavikrds [6]

Answer:

A Winter storm usually does not come with a warm fronthey warm front.

Explanation:

When the front passes over an area it means to change in the weather. Many. Many friends cause weather events such as rain thunderstorms gusty winds tornadoes and hurricanes.

5 0
2 years ago
Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
mylen [45]

Answer:

1.06  V  

Explanation:

The standard reduction potentials are:

Ag^+/Ag     E° =  0.7996 V  

Ni^2+/Ni     E° = -0.257   V

The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e-                     E° = 0.257   V

<u>2Ag^+ 2e- → 2Ag               </u>    <u>E° = 0.7996 V </u>

Ni + 2Ag^+ → Ni^2+ + 2Ag     E° = 1.0566  V

To three significant figures, the standard potential for the cell is 1.06 V .

8 0
4 years ago
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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