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notka56 [123]
3 years ago
13

A woman runs a marathon in 90 ºf weather and replaces all volume lost in sweat by drinking distilled water. After the marathon,

she will have:.
Chemistry
1 answer:
gayaneshka [121]3 years ago
6 0

Answer:

Total body water (TBW) will be unchanged because the woman replaced all volume lost in sweat by drinking water. Hematocrit will be increased because of the shift of water from ECF to ICF and the shift of water into the red blood cells (RBCs), which causes their volume to increase.

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In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

4 0
3 years ago
Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2
Salsk061 [2.6K]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

2C(s)+2H_2(g)\rightarrow C_2H_4(g) \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.7kJ    ( × 2)

(2) H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_2=-285.9kJ     ( × 2)

(3) 2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)    \Delta H_3=-1411kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is 51.8 kJ.

6 0
3 years ago
Reactants are substances that?
RSB [31]
A substance that undergoes change during a reaction, usually from coming in contact with another substance
4 0
3 years ago
Which of the following electron configurations are written incorrectly?
Lynna [10]

Answer:

The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.

Explanation:

The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.

There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.

All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".

In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.

The other options are correctly written.

3 0
3 years ago
The units of work are force and energy im not sure if im right
Andrej [43]
The unit for work is ENERGY
6 0
3 years ago
Read 2 more answers
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