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spin [16.1K]
3 years ago
7

Plz help me i have to give this by today pls...........

Chemistry
1 answer:
vfiekz [6]3 years ago
7 0
I need a a picture or the question
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How many liters of softened water, containing a sodium concentration of 5.2×10−2 % sodium by mass, have to be consumed to exceed
german

Answer:

4423.08 L

Explanation:

First we need to know the concentration of sodium per litre (as the density is 1 g/L, 1 g of water is the same that 1 mL of water) of softened water by dividing the percentage in 100:

5.2x10^{-4}g/L

With the value of concentration, we just need to to divide the recommended value of the FDA, that is 2300 milligrams per day, into the concentration that we have, therefore:

2300mg=2.3g\\\\\frac{2.3g}{5.2x10^{-4}g/L} =4423.08 L

8 0
3 years ago
Place the following in order of increasing molar entropy at 298 K.
Genrish500 [490]

Answer:

SO < CO2 < C3H8

Explanation:

Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.

The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.

C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.

3 0
3 years ago
What is the mass of 4.25 liters of N2 gas at STP?
sukhopar [10]

We know that 1 mole of anything has a volume of 22.4 liters.

And 1 mole of N2 gas has a weight of 28 grams.

Comparing them , we get the answer as 5.31

7 0
3 years ago
Read 2 more answers
Electrons =28 so what is the element
maksim [4K]

Answer:

Nickel is the answer

Explanation:

Lol that's easy the atomic number is the same number as the number of electrons.

8 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
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