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3 years ago

8

What is the ionic bonding of NaCl

1 answer:

madam [21]3 years ago

8 0

Na has a one proton over than electrons.so Na makes a Na+
Cl has a one electron overthan protons.so Cl makes a Cl-

proton has a + affectation
electron has a - affectation ..

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Where are angora goats found

jok3333 [9.3K]

Answer:

These thick-haired goats originate in Turkey

Explanation:

and are named after the country's capital, Ankara, historically known as Angora.

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3 years ago

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A car is traveling 71.4 kilometers per hour. What is its speed in miles per minute? One kilometer = 0.62 miles.

Nimfa-mama [501]

$71.4 \times 0.62 = 44.268$

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3 years ago

BRO PLSSSS HELP

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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

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3 years ago

Rank the following solutions in order of increasing acidity (least acidic to most acidic): I: [H3O+] = 1x10−5 II: [OH−] = 1x10−1

kozerog [31]

Answer: option C) II < III < I

i.e [OH−] < [H3O+] < I

Explanation:

First, obtain the pH value of I and II, then compare both with III.

For I

Recall that pH = -log (H+)

So pH3O = -log (H3O+)

= - log (1x10−5)

= 4

For II

pOH = - log(OH-)

= - log(1x10−10)

= 9

For III

pH = 6

Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4

Thus, the following solutions from least acidic to most acidic is II < III < I

6 0

3 years ago