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2 years ago

PLEASE HELP ASAP

Chemistry

2 answers:

Digiron [165]2 years ago

5 0

Answer:

Limestone

Explanation:

Limestone is required , hope it helps :)

DedPeter [7]2 years ago

5 0

A is the answer

Answer: Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves

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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)

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

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

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3 years ago

Formed when chemicals in the air get into rain and up the acidity levels

snow_lady [41]

Acid Rain is formed when chemicals in the air get into rain and up the acidity levels!!!!!!

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4 0

3 years ago

What is the total mass of Hydrogen in each of the molecules?

frutty [35]

Explanation:

a) In 1 mole of methane there are 4 moles of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of methane = 4 × 1 g = 4 g

b) In 1 mole of chloroform there are 1 mole of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of methane = 1× 1 g = 1 g

c) In 1 mole of $C_{12}H_{10}O_{16}$ there are 10 moles of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of $C_{12}H_{10}O_{16}$ = 10 × 1 g = 10 g

d)In 1 mole of $CH_3CH_2CH_2CH_2CH_3$ there are 12 moles of hydrogen atom.

$CH_3CH_2CH_2CH_2CH_3=C_5H_{12}$

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of $CH_3CH_2CH_2CH_2CH_3$ = 12 × 1 g = 12 g

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4 years ago

How many molecules of H2O are equivalent to 97.2g H2O

IgorLugansk [536]

como se denomina el proceso utilizado para descomponer el agua

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3 years ago

Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0g of potassium and

Leona [35]

Answer:

79.43kg

Explanation:

To adequately solve this problem, we should know the law of constant composition. This is a pointer to the fact that no matter the type of sample, the percentage compositions of potassium and iodine still remains the same. We can use these masses to get a formula and we would know the percentage compositions in whatever mass we are dealing with.

First of all, we add the masses in the first sample. 13 + 42.3 = 55.3

Hence, the percentage composition of the potassium is 13/55.3 * 100 = 23.51%

The percentage composition of the iodine is = 100 - 23.51 = 76.49%

Now, we need to get the formula of the compound. We can get this by dividing the percentage compositions with the atomic masses. The atomic mass of potassium and iodine is 39 and 127 respectively.

Potassium = 23.51/39 =0.603

Iodine = 76.49/127 = 0.602

We then divide by the smaller value to get the formula and this shoes our formula is KI

We can see they have a ratio of 1 to 1, meaning one atom of potassium to one atom of iodine. This further confirms the percentage compositions of 23.5 to 76.5

Now to get the mass of iodine yielded, let us say the mass is xkg

This means x/(x + 24.4) * 100 = 76.5

100x = 76.5( x + 24.4)

100x = 76.5x + 1866.6

100x - 76.5x = 1866.6

23.5x = 1866.6

x = 1866.6/23.5 = 79.43kg

4 0

3 years ago