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3 years ago

What is the lengthening of a muscle without damage? And what is the shortening of a muscle?

Chemistry

1 answer:

siniylev [52]3 years ago

5 0

Answer:

Stretching is the lengthening and contraction is the shortening

Explanation:

Unless it's asking for eccentric as the stretching and concentric as the shortening

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Which element has a +2 charge and has a configuration of 1s22s22p3? A) B B) F

PSYCHO15rus [73]

Fluorine(F)

Explanation:-

Fluorine has Z=9
+2 charge means 2 electrons donated (7left only)

Lets look at Electronic configuration

$\\ \sf\longmapsto 1s^22s^22p^3$

6 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Coke is an impure form of carbon that is often used in the in-dustrial production of metals from their oxides. if a sample of co

olasank [31]

The reaction equation is:
2CuO(s) + C(s) → 2Cu(s) + CO₂(g)

First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10⁵ g

We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles

Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required

The mass of carbon is then:
5,717 x 12 = 68,604 grams

The mass of coke is:
68,604 / 0.95 = 72,214 g

The mass of coke required is 7.22 x 10⁴ grams

5 0

3 years ago

Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in

AysviL [449]

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = $m_{water}\times C_{water}\times \Delta T_{water}$

where m represents mass, C represents specific heat and $\Delta T$ represents change in temperature.

Here $m_{water}=75.0$ g , $C_{water}=4.184J/(g.^{0}\textrm{C})$ and $\Delta T$ = (final temperature - initial temperature) = (29.5-21.2) $^{0}\textrm{C}$ = 8.3 $^{0}\textrm{C}$