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3 years ago

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3. A scientific investigation typically begins when a scientist makes observations that lead to the formation of a (n) _________

_________________?
A.law
B.theory
C.hypothesis
D.none of the above

1 answer:

Artist 52 [7]3 years ago

8 0

Answer:

C. Hypothesis

Explanation:

A hypothesis is usually the start of a scientific investigation.

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How many grams of water are produced if we react 3 moles of hydrogen with 3 moles of oxygen?

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Answer:

About 60 grams

Explanation:

The balanced equation of the reaction is given as;

2H2(g) + O2(g) → 2H2O(l)

From the equation;

2 mol of H2 reacts with 1 mol of O2

3 mol of H2 would require 3/2 mol of O2 (considering the 2:1 ratio)

The limiting reactant is H2 as it would be used up before H2. It determines the amount of product that would be formed.

2 mol of H2 produces 2 mol of H2O

3 mol of H2 would produce 3 mol of H2O (Considering the 1 : 1 ratio)

Converting moles to mass;

Mass = Molar mass * Number of moles

Mass = 18 * 3 = 54 g

The correct option is; About 60 grams

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2 years ago

In the periodic table of elements,all <br>the elements in group 16 have the same number of what

Maksim231197 [3]

Elements in the same group share the same number of valence electrons.

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3 years ago

Read 2 more answers

A student chromatographs a mixture and after developing the spots with a suitable reagent he observes the following

Anika [276]

Rf value is the ratio of the distance traveled by the solute to that of the solvent front on the paper used in chromatographic separation.

From the image it is clear the distance traveled by solvent front = 7.3 cm

Distance traveled by the component -1 of the mixture = 1.4 cm

Distance traveled by the component -2 of the mixture = 3.0 cm

Distance traveled by the component -3 of the mixture = 4.5 cm

Distance traveled by the component -4 of the mixture = 6.5 cm

Rf value of component-1 = $\frac{1.4 cm}{7.3 cm} =0.192$

Rf value of component-2 = $\frac{3.0 cm}{7.3 cm} =0.410$

Rf value of component-3 = $\frac{4.5 cm}{7.3 cm} =0.616$

Rf value of component-4 = $\frac{6.5 cm}{7.3 cm} =0.890$

b) Samples can be separated from a mixture using chromatography as the relative affinities for the compounds towards the paper (stationary phase) and the solvent(mobile phase) are different. Each component spends different amounts of time on the stationary phase depending on it chemical nature. So, the components in a mixture can be separated based on their polarities and relative degrees of adsorption on the stationary phase.

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3 years ago

Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s

disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R $HSO_4^-$ ⇄ $H^+ + SO_4^2^-$

I $0.14$

C $- x$ $+x$ $+x$

E $0.14-x$ $x$ $x$

$K_a = 1.3$ × $10^-^2$ for $HSO^-_4$ . As a result,

$\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a$

$K_a$ is large. It is no longer valid to approximate that $[HSO^-_4]$ at equilibrium is the same as its initial value.

$\frac{x^2}{0.14-y} = 1.3 * 10^-^2$

$x^2+1.3*10^-^2x - 0.14$ × $1.3$ × $10^-^2= 0$

Solving the quadratic equation for $x , x \geq 0$ since $x$ represents a concentration;

$x=0.0366538$

Then, round the results to 2 significant figure;

$[SO_4^2^-] = x = 0.037 mol. L ^-^1$
$[H^+] = x = 0.037 mol. L ^-^1$
$[HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1$

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

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1 year ago