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4 years ago

Determine whether the statement is true or false, and why. "If a theory becomes supported by evidence, it can become a law."

1 answer:

5 0

Answer: D. False, it should read, "A theory and a law are already both supported by evidence and are equal, but they have different functions."

Explanation:

A "Theory" and a'' Law" in scientific method are already both supported by evidence but they have different function,

The function of " Theory" in scientific method is to give an explanation to the observations and findings gathered during the process. It can be modified, improved or even rejected as more information is being gathered.

While a '' Law '' tends to describe an observed event in nature that is true whenever it is put to test. It does not explain why the occurence or event is true.

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Determine the name of the following compound Fe(OH) 3 , Provide explanation of how you determined the name, including the type o

stepan [7]

Answer:

1. Iron(iii) hydroxide

2. It is a base

Explanation:

To determine the name of the compound, we must find the oxidation number Fe in the compound since the oxidation number of O is —2 in all its compound except for peroxide where it —1 and the oxidation number of Hydrogen is always +1 in all its compounds except for hydrides where it is —1. The oxidation number of Fe can calculated as follows:

Fe(OH)3 = 0

O = —2

H = +1

Fe =?

Fe + 3( —2 + 1) = 0

Fe + 3(—1) = 0

Fe —3 = 0

Fe = 3

The oxidation number of Fe in Fe(OH)3 is 3.

Therefore, the name of Fe(OH)3 is Iron(iii) hydroxide

Fe(OH)3 is a base since it contains the OH group

7 0

4 years ago

In organisms whose internal body temperature has dropped, which of the following responses is likely to result?

vichka [17]

C, Shivering. The only reason im writing this is because i need 20 characters

3 0

3 years ago

Write balanced molecular, ionic, and net ionic equations for the reaction occurs when (a) a BaCl2 (Barium Chloride) is mixed wit

lubasha [3.4K]

A.) We write the molecular equation for the chemical reaction as:

BaCl2(aq)+Na2SO4(aq)→BaSO4(s)⏐+2NaCl(aq)

Alternatively, we could write the net ionic equation:

Ba2+ + SO2^−4→ BaSO4(s)⏐

B.)We write the molecular equation for the chemical reaction as:

Ba(OH)^2 + 2HBr → BaBr2 + 2H2O

The ionic equation can be written as;

Ba^2+ + 2OH^- + 2H^+ + 2Br^- →

Ba^2+ + 2Br^- + 2H2O

the net ionic equation can be written as;

2OH^- + 2H^+ → 2H2O

Alternatively, we could write the net ionic equation

OH^-(aq) + H^+(aq) → H2O(l) net ionic equation.

C.) We write the molecular equation for the chemical reaction as:

2 HNO2(aq) + Ba(OH)2(aq) 

→ Ba(NO2)2(aq) + 2 H2O(l)

Net ionic equation

HNO2(aq) + OH-(aq) → NO2-(aq) + H2O(l)

Alternatively, we could write the net ionic equation as:

2H^+ + 2OH → 2H2O

8 0

4 years ago

0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.

max2010maxim [7]

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

7 0

3 years ago

g At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 (g) Br 2 (g)

oksano4ka [1.4K]

Answer: At equilibrium , there are 0.274 moles of $Br_2$

Explanation:

Moles of $H_2$ = 0.682 mole

Moles of $Br_2$ = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of $H_2$ = $\frac{0.682}{2.00}=0.341 M$

Initial concentration of $Br_2$ = $\frac{0.440}{2.00}=0.220 M$

Equilibrium concentration of $H_2$ = $\frac{0.516}{2.00}=0.258 M$

The given balanced equilibrium reaction is,

$H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)$

Initial conc. 0.341 M 0.220 M 0 M

At eqm. conc. (0.341-x) M (0.220-x) M (2x) M

Given : (0.341-x) M = 0.258 M

x= 0.083 M

Thus equilibrium concentartion of $Br_2$ = (0.220-0.083) M = 0.137 M

Thus moles of $Br_2$ at equilibrium = $0.137M\times 2.00L=0.274mol$

At equilibrium , there are 0.274 moles of $Br_2$

8 0

3 years ago