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3 years ago

Name the ionic compound represented by the following chemical formula: CuO

Chemistry

1 answer:

Illusion [34]3 years ago

6 0

Answer:

Cupric oxide, or copper (II) oxide,

Explanation:

Correct me if im wrong tnx:<

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The pressure of a gas is 750 torr when its volume is 400. ml. Calculate the

ololo11 [35]

Answer:

a. 0.66 atm

Explanation:

Pi = 750 torr = 0.99 atm. Pf = ?

Vi = 400 ml Vf = 600ml

Using PiVi = PfVf

Then 400ml × 0.99atm = Pfatm × 600ml

Solving for Pf, we have,

Pf = (400ml × 0.99atm)/600ml

Pf = 0.66atm

7 0

3 years ago

An empty graduated cylinder weighs 26.5 grams and when it is filled with an unknown liquid to the 45.8 ml mark the cylinder and

Softa [21]

Answer:

24.2 is the density of the liquid

7 0

3 years ago

Andi performs the calculation that is shown below.

rodikova [14]

Answer: 3.59

Explanation:

(2.06)(1.743)(1.00)

2.06 × 1.743 × 1.00

= 3.59058

Two of the multiplied digits are represented in 3 significant figures. Therefore, for correct representation, the result of the product should be written to three significant figures.

3.59058 to 3 significant figures:

First three digits = 3.59

Fourth digit '0' is less than 5, and thus rounded to 0 with other succeeding digits

Therefore, (2.06)(1.743)(1.00) to 3 significant figures equals :

3.59

3 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Leni [432]

Answer:

S(s) + O2(g) --> SO2(g)

Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

Explanation:

The reaction is given as;

Sulfur + oxygen --> Sulphur dioxide

Sulphur = S

Oxygen = O2

Sulfur dioxide = SO2

So we have;

S(s) + O2(g) --> SO2(g)

The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

5 0

3 years ago

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