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amm1812
3 years ago
14

PLS HELP!!

Chemistry
1 answer:
zimovet [89]3 years ago
3 0

Answer:

if its a multiple question answer  its hydroden  and carbon

if not its carbon

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Warm Up: Compete this conversion. Put just the number below *
Gwar [14]

Answer:

4 1/2

Explanation:

Use a ratio to find your answer

 4          6

-----  =  -------

 3           x

Cross multiply to solve for x.

4x = 18

x = 18/4

x = 4 2/4  which is the same as 4 1/2

5 0
2 years ago
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Blizzard [7]

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

8 0
3 years ago
Which method is suitable for collection a a<br>carbon dioxide.Explain​
mel-nik [20]

Answer:

Carbon dioxide can be collected over water. Carbon dioxide is slightly soluble in water and denser than air, so another way to collect it is in a dry, upright gas jar.

Explanation:

8 0
3 years ago
What is the answers and pls show work if possible!!
taurus [48]

D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

Diameter = 348 pm


46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold


4 0
3 years ago
A 687.80-g sample of a noble is held in a 15 L cylinder tank at 3800 torr and a temperature of 22 degrees C. Identify the gas.
BARSIC [14]

Answer:.......................

8 0
3 years ago
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