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3 years ago

The pH of an unknown strong acid is 2. What is the hydronium ion concentration?

1 answer:

8 0

Answer: Finding the [H3O+] and pH of Strong and Weak Acid Solutions The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium. hydronium ion, H3O+, 1.0, 0.00, H2O, 1.0×10−14, 14.00.

Explanation:The hydrogen ion in aqueous solution is no more than a proton, a bare ... the interaction between H+ and H2O .

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Two things plants need to have in order to stay in orbit and rotate is speed and less velocity.

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3 years ago

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3 years ago

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

This value corresponds to iron. Thus answer is B.

3 0

3 years ago

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.