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Kazeer [188]
4 years ago
7

What do dinking objects have in common

Chemistry
1 answer:
Viefleur [7K]4 years ago
5 0
They all have densities greater than the density of the fluid in which they are<span>sinking. The mass of the displaced liquid is less than the mass of the sinking body.</span>
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What temperature change in Celsius degrees is produced when 800. J is absorbed by 100. G of water?
g100num [7]

Answer: The final temperature is equal to

45 Celsius

Explanation:

Explanation:

We have the amount of energy gain

Q

=

m

⋅

c

⋅

Δ

T

=

m

c

Δ

t

where

c

=

4.184

J

/

g

.

C

is the specific heat of water,

m

is the mass of water

⇒

840

=

10

x

4.184

⋅

(

t

−

25

)

t

=

840

10

x

4.184

+

25

=

45

i.e.

45

∘

C

6 0
4 years ago
In order for an object
yarga [219]

Answer:

uho9u8

Explanation:

5 0
3 years ago
A NaOH solution is standardized using the monoprotic primary standard potassium hydrogen phthalate, KHP (204.22 g/mol.) If 0.698
kakasveta [241]

Answer:

0.07789 M is the sodium hydroxide concentration.

Explanation:

Mass of potassium hydrogen phthalate = 0.6986 g

Molar mass of potassium hydrogen phthalate = 204.22 g/mol

Moles of  potassium hydrogen phthalate = \frac{0.6986 g}{204.22 g/mol}=0.003421 mol

NaOH+KHP\rightarrow NaKP+H_2O

According to reaction , 1 mole og KHp reactswith 1 mole of NaOH , then 0.003421 moles of KHp will react with :

\frac{1}{1}\times 0.003421 mol=0.003421 mol

Moles of NaOH = 0.003421 mole

Volume of NaOH solution = 43.92 ml = 0.04392 L ( 1 mL = 0.001L)

Concentration=\frac{Moles}{Volume(L)}

Concentration of NaOH :

\frac{0.003421 mol}{0.04392 L}=0.07789 M

0.07789 M is the sodium hydroxide concentration.

5 0
3 years ago
Calculate the density of a liquid if 58.9 ml of it has a mass of 46.08 g. answer in units of g/ml.
alexira [117]
Density of a solution is mass of solution per unit volume
Density = mass/volume
mass of solution is 46.08 g
volume of solution is 58.9 mL 
since mass and volume is known, density can be calculated
density = 46.08 g / 58.9 mL = 0.78 g/mL 
8 0
3 years ago
Consider a galvanic cell in which Al 3 + Al3+ is reduced to elemental aluminum and magnesium metal is oxidized to Mg 2 + Mg2+ .
olga nikolaevna [1]

Answer:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

Explanation:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

Magnesium is more electro positive than aluminum hence it functions as the anode. Six electrons are lost/gained in the redox process as shown in the oxidation and reduction half reaction equations above. Magnesium is oxidized to magnesium ion while aluminum is reduced to elemental aluminum.

5 0
4 years ago
Read 2 more answers
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