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Sever21 [200]
3 years ago
12

Using your knowledge of symbols, determine which compound listed must contain both copper and oxygen.

Chemistry
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

Malachite

Explanation:

Malachite is the only listed compound that must contain copper and oxygen.

Copper and oxygen are both elements found on the periodic table. They have the following symbols;

         Copper  = Cu

         Oxygen  = O

From the given choices, only option 1 has the symbol Cu and O.

So only malachite contains both copper and oxygen.

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Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.
velikii [3]
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
6 0
3 years ago
MARK BRAINLIEST AND 14 POINTS
LekaFEV [45]

Answer:

She could prove that it is a combination of substances by looking for a change in color, or the formation of bubbles. She could also try to pull the combination apart by physical means alone.

Explanation:

4 0
3 years ago
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Draw the structure of 5 bromo 1,1,1-trichloro-5 ethyl <br> 2,3 di methyl heptane
andre [41]

5 bromo 1.1.1 -trichloro -5 ethyl 2.3 dimethyl heptane

5 0
3 years ago
PLEASE HELP ME ASAPPPPP What is the density of a piece of cardboard that has a mass of 250 g and volume of 46 mL? *
Roman55 [17]

Answer:

The answer is

<h2>5.43 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 250 g

volume = 46 mL

The density is

density =  \frac{250}{46}  \\  = 5.4347826...

We have the final answer as

<h3>5.43 g/mL</h3>

Hope this helps you

6 0
3 years ago
7. How many moles of argon are there in 20.0 L, at 25 degrees Celsius and 96.8 kPa?
suter [353]
<h3>Answer:</h3>

              0.8133 mol

<h3>Solution:</h3>

Data Given:

                 Moles  =  n  =  ??

                 Temperature  =  T  =  25 °C + 273.15  =  298.15 K

                  Pressure  =  P  =  96.8 kPa  =  0.955 atm

                  Volume  =  V  =  20.0 L

Formula Used:

Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for n,

                  n  =  P V / R T

Putting Values,

                  n  =  (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)

                 n  =  0.8133 mol

4 0
3 years ago
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