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Leokris [45]
3 years ago
11

The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium

could be obtained from 9.84x10^8 cubic meter of seawater. Assuming the density of sea water is 1.0g/mL.
Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
4 0

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

 \frac{9 .84x10^8}{1000000}      = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = \frac{mass}{volume}

mass = density x volume

mass = 1 x 984

         = 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

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How much heat is needed to vaporize 10.0 grams of water?
Furkat [3]
10*2260=22600j=22.6kJ
8 0
3 years ago
What is the specific heat (J/g C)of a metal who’s temp increases by 3.0 C when 17.5 g of metal was heated with 38.5 J
disa [49]

Answer:

0.733 J/g °C

Explanation:

<u>Step 1 : Write formule of specific heat</u>

<u />

Q = mc Δ T

with Q = heat transfer (J)

with m = mass of the substance

with c = specific heat ⇒ depends on material and phase ( J/g °C)

Δ T = Change in temperature

<u />

in this case :

Q = 38.5 J

<u>m = 17.5g</u>

<u>c= has to be determined</u>

<u>Δ T = 3 (°C)</u>

<u />

<u>Step 2:  Calculating specific heat</u>

⇒ via the formule Q = mc Δ T

38.5 J = 17.5g * c * 3

38.5 = 52.5 *c

c = 38.5 / 52.5

c = 0.733 J/g °C ⇒ 0.733 is reported to three significant digits due to the precision.

The specific heat of this metal is 0.733 J/g °C

8 0
3 years ago
Help I need to show work!
lubasha [3.4K]

Answer:

a. Combustion.

b. m_C=3.057gC

c. m_H=3.67gH

Explanation:

Hello.

a. In this case, given the reaction by which propane is converted into carbon dioxide and water:

C_3H_8+5O_2 \rightarrow 3CO_2+4H_2O

It is known as a combustion reaction since it is about a fuel (here propane) which is burnt by the oxygen contained in the air (21%).

b. Since the carbon dioxide contains all the carbon in the products based on the law of conservation of mass, the yielded grams are computed via a mole mass relationship:

m_C=11.21gCO_2*\frac{12gC}{44gCO_2}=3.057gC

Since 44 grams of carbon dioxide contain 12 grams of carbon.

c. As well as b., all the hydrogen is  given off in form of water, thus, the required mass turns out:

m_H=33.0gH_2O*\frac{2gH}{18gH_2O}=3.67gH

Since 18 grams of water contain 2 grams of hydrogen.

Best regards.

6 0
3 years ago
A solution of sodium iodide is added to a solution of potassium nitrate to make a potassium iodide precipitate and a sodium nitr
stich3 [128]

Answer:

NaNO3

Explanation:

Sorry If its incorrect

4 0
3 years ago
In the following reaction, when a surplus of hydrazine (N2H4) reacts with 20 liters of hydrogen peroxide (H2O2), how many liters
Ghella [55]
The answer is 10 liters.
5 0
3 years ago
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