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stepladder [879]
1 year ago
11

True or false? the iron recommendation for girls exceeds that of boys during adolescence.

Chemistry
1 answer:
sashaice [31]1 year ago
4 0

It is true that the iron recommendation for girls exceeds that of boys during adolescence.

During an adolescence, girls and boys require iron for a large growth spurt and the gain of adult phenotypes and biologic rhythms.

In this period of the life, iron recommendation increase in both girls and boys, because of the increase in lean body mass, the expansion of the total blood volume, the increase and start of menstruation at girls.

Iron is essential for oxygen transport, red blood cell creation, cognitive performance and immunological function.

The overall iron requirements for girls are up to twice as boys.

More about adolescence: brainly.com/question/13528489

#SPJ4

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Answer:

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Explanation:

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3 0
3 years ago
1.00 L of gas at a standard temperature and pressure is compressed to 473mL. What is the new pressure of the gas?
Nadusha1986 [10]

Answer:

The pressure of the gas is 2.11 atm.

Explanation:

From the given,

V_{1}=1.00\,L

P_{1}=1\,atm

P_{2}=?

V_{2}=473\,ml=0.473\,L

P_{1}V_{1}=P_{2}V_{2}

(1)(1)=(0.473)(P_{2})

P_{2}=\frac{(1)(1)}{0.473}=2.11

Therefore, The pressure of the gas is 2.11 atm.

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3 years ago
What are quantum numbers?
kenny6666 [7]

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3 0
3 years ago
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What structure is found in some but not all plant cells
AlekseyPX
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5 0
3 years ago
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From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

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

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
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