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3 years ago

I don't know how to do any of these.plz help.its a study guide for my chemistry exam tomorrow ✌

Chemistry

1 answer:

Vladimir [108]3 years ago

8 0

6.02 x 1023 atoms weigh out 63.55 grams copper.

No. of Molecules in water = 3.5mole x (6.02 x 10^23) molecules/mole = 2.107 x 10^24 molecules of H2O

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Why are humans conductive

Andrews [41]

Look it up to find the answer . they will tell u all bout humus

4 0

3 years ago

Now suppose, instead, that 5.678 g of a volatile solute is dissolved in 150.0 g of water. This solute also does not react with w

n200080 [17]

Answer:

59.9 g/mol is the molar mass for the solute

Explanation:

Lowering vapor pressure → ΔP = P° . Xm

P° → Vapor pressure of pure solvent

ΔP = P° - Vapor pressure of solution

Xm = Mole fraction of solute

17.54 Torr - 17.344 Torr = 17.54 Torr . Xm

0.196 Torr / 17.54 Torr = Xm → 0.0112

These are the moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

We determine the moles of solvent → 150 g . 1mol/ 18 g = 8.33 moles

Now we can make this equation:

0.0112 = Moles of solute / Moles of solute + 8.33 mol

0.0112 Moles of solute + 0.0933 = Moles of solute

0.0933 = Moles of solute - 0.0112 Moles of solute

0.0933 = 0.9888 moles of solute → 0.0933 / 0.9888 = 0.0947 moles

Finally we can determine the molar mass (mol/g)

5.678 g / 0.0947 mol = 59.9 g/mol

4 0

3 years ago

If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g･°C) that can be heated from 22.5°C to

inysia [295]

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

Specific heat,

C = 0.45 J/g･°C

Heat involved,

q = 801 J

Temperature,

ΔT = 120.0°C-22.5°C

= 97.5°C

As we know,

⇒ $C = \frac{q}{m \Delta T}$

On substituting the given values, we get

⇒ $0.45=\frac{801}{m(97.5)}$

⇒ $m = 18.25 \ g$

3 0

4 years ago

Describe gas production

kaheart [24]

Answer:

it natural gas or fossil fuels

5 0

3 years ago

For a solution equimolar in HCN and NaCN, which statement is false? a. [H+] is larger than it would be if only the HCN were in s

balandron [24]

Answer:

option (d) is false.

Explanation:

Acid dissociation equilibrium of HCN is represented as-

$HCN\rightleftharpoons H^{+}+CN^{-}$

Acid dissociation constant, $K_{a}$ , is represented as-

$K_{a}=\frac{[H^{+}][CN^{-}]}{[HCN]}$

where species inside third bracket represents equilibrium concentrations of respective species

So, evidently, presence of excess $CN^{-}$ (or NaCN) in solution will combine with $H^{+}$ to produce HCN. Hence $H^{+}$ will be larger that it would be if only the HCN solution were present.

According to Le-chatlier principle, addition of HCN will shift equilibrium towards right and addition of NaCN will shift equilibrium towards left to keep constant $K_{a}$ value at a particular temperature.

NaOH gives acid-base reaction with HCN to produce NaCN and water. So, addition of NaOH will increase concentration of $CN^{-}$ and decrease concentration of HCN

6 0

4 years ago