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3 years ago

Heat and pressure deep beneath Earths Surface any rock into

Chemistry

1 answer:

MrRa [10]3 years ago

5 0

METAMORPHIC ROCK is if its correct

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Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C,

AlladinOne [14]

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

7 0

3 years ago

What is the net ionic equation for this reaction? <br><br> FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl

jok3333 [9.3K]

Answer:

2.6299

its easy

8 0

3 years ago

Christy is a forensic scientist. She is given

Nitella [24]

Answer:

The white powder is not sugar, so Christy's hypothesis is incorrect . Christy's next step should be to identify substances that have the measured molecular mass.

Explanation:

6 0

4 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

What is the percent of composition by mass of hydrogen in NH4HCO3 (gram-formualmass=79 grams/mole)?

musickatia [10]

5/79=0.063
0.063 x 100 = 6.3%

5 0

3 years ago