All categories

3 years ago

Please help me please

Chemistry

2 answers:

AnnZ [28]3 years ago

8 0

Answer:

20 is b and 21 is a.

Ludmilka [50]3 years ago

7 0

Answer:

Explanation:

i read this once and remember,trust me!

You might be interested in

An unknown solution has a pH of 2. How would you classify this solution?

devlian [24]

Acidic because anything over a ph of 7 would be basic but anything below 7 would be acid and if it has a ph of 7 it would be neutral....Hope I helped

7 0

3 years ago

Which is the correct set of products for the acid-base neutralization reaction between sulfuric acid (H2SO4) and potassium hydro

Elis [28]

Answer:

H2O and K2SO4

Explanation:

3 0

3 years ago

How do cheetos store energy?

just olya [345]

Answer:

Calories

Explanation:

they cheeto substance is compared in various ways but our body consoles it by its calories which is a unit of energy as it give us the energy to keep our body warm

5 0

3 years ago

N each reaction box, place the best reagent or reactant from the list below. reagents may be used more than once or not at all.

Ne4ueva [31]

Answer:

See explaination

Explanation:

See attachment for the drawing of the intermediate products b and c (both are neutral; omit byproducts).

7 0

3 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

3 years ago