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3 years ago

Copper is an example of a(n) __________.

Chemistry

2 answers:

Tpy6a [65]3 years ago

8 0

Hi, the answer is D. Element.

navik [9.2K]3 years ago

7 0

The answer is D. element

It's not a mixture, just a pure element. Copper (Cu) is found on the Periodic Table of the Elements, where only ELEMENTS are organized.

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A pump contains 0.5 L of air at 203kPa. You draw back on the piston of the pump until the pressure reads 25.4kPa. What is the vo

densk [106]

<h3> Answer;</h3>

= 4.0 L

<h3>Explanation;</h3>

Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.

Therefore; Volume α 1/pressure

Mathematically; V α 1/P

V = kP, where k is a constant;

P1V1 = P2V2

V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa

V2 = (0.5 × 203 )/25.4 

 = 3.996 

 ≈ 4.0 L

6 0

3 years ago

In three different experiments, the following results were obtained for the reaction

faust18 [17]

From the given observations,
You can see that as the concentration is doubled, half-life is halved.

That is,half-life is inversely proportional to concentration

As t( half-life) ~ 1/a^(n-1)

For this case n = 2,second order reaction.

R = k X a^n

Using the above formula you will get the rate and rate constant.

5 0

3 years ago

Potassium hydroxide molar mass

Mnenie [13.5K]

The answer I will give is an approximate number, but it will be close. The problem is that every periodic table is slightly different. Some round as I will, and some carry the mass of the elements out to 3 decimal places. You should go back and put the correct numbers in to the mass that I use.

Formula

KOH

Givens

K = 39
O = 16
H = 1

Solution

Molar Mass = 39 + 16 + 1

Molar Mass = 56 grams / mole

6 0

3 years ago

Mn(OH)2(s) + MnO4(aq) → MnO42–(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what

PilotLPTM [1.2K]

Answer:

Hi, the given equation has some missing parts. Actual equation is- ' $Mn(OH)_{2}(s)+MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$ '

balanced equation: $Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Explanation:

$Mn(OH)_{2}(s)\rightarrow MnO_{4}^{2-}(aq.)$

Balance O and H in basic medium: $Mn(OH)_{2}(s)+6OH^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Balance charge: $Mn(OH)_{2}(s)+6OH^{-}(aq.)-4e^{-}\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$ ........(1)

$MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$

Balance charge: $MnO_{4}^{-}(aq.)+e^{-}\rightarrow MnO_{4}^{2-}(aq.)$ .....(2)

$[equation(2)\times 4]+[equation (1)]:$

$Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

$OH^{-}(aq.)$ is present on the left hand side of balanced equation and it's coefficient is 6

6 0

2 years ago

In effect of concentration on rate,why is it important to

MrRissso [65]

Explanation:

If we change the concentration of Potassium since we take KI solution, concentration of I- changes, then the rate changes accordingly

Rate = k[H2O2 ]^a[I-]^b [H+]^c

The concentrations of I- and H+ are held constant in the procedure

thus to study the rate of the reaction, concentration of KI solution has to be constant only the peroxide solution varies.

7 0

3 years ago