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OleMash [197]
3 years ago
9

What element is found in liver and is needed to prevent anemia or “tired” blood?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

Iron (Fe)

Explanation:

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What’s the difference between a physical and chemical property
irga5000 [103]

Answer:

One is hands on and one is watching

Explanation:

I just know my science

3 0
3 years ago
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Dovator [93]

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

3 0
3 years ago
A certain sample of coal contains 1.60 percent sulfur by mass. when the coal is burned, the sulfur is converted to sulfur dioxid
alina1380 [7]
<span>1.40 x 10^5 kilograms of calcium oxide The reaction looks like SO2 + CaO => CaSO3 First, determine the mass of sulfur in the coal 5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4 Now lookup the atomic weights of Sulfur, Calcium, and Oxygen. Sulfur = 32.065 Calcium = 40.078 Oxygen = 15.999 Calculate the molar mass of CaO CaO = 40.078 + 15.999 = 56.077 Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight. 8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass 2.49 x 10^3 * 56.077 = 1.40 x 10^5 So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
5 0
3 years ago
An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the
victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

  • To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:  

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

We know that:

K_a\text{ for HF}=6.8\times 10^{-6}

K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

So, the calculated K_a is approximately equal to the K_a of HF

Hence, the unknown salt is NaF

6 0
3 years ago
Does the Ba(OH)2 solution contain ions?
LUCKY_DIMON [66]
Yes. It does contain Ions. 
4 0
2 years ago
Read 2 more answers
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