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3 years ago

What are the 5 steps of conversion

Chemistry

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer:

Step 1 – Research & profiling.

Step 2 – Hypothesis & variables.

Step 3 – Design.

Step 4 – Implementation & testing.

Step 5 – Improvements.

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A bottle contains 3.100 ml of a liquid. the total mass of the bottle and the liquid together is 6.300 g. the mass of the empty b

Olenka [21]

The answer is <span>a. 0.665 g/m</span>

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4 years ago

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While preforming an experiment involving colors and heat,Susan noticed that the black paper was warmer than the white piece of p

Firdavs [7]

Answer:

Blackbody radiation

Explanation:

This phenomenon described by Susan is know as black body radiation. Different materials by color reacts to incident radiation on them in diverse manner.

A white surface will reflect all the wavelength of light incident on it.
A black surface will absorb all the wavelength of light on it.

This absorption causes the black surface to be hotter compared to other surfaces.

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3 years ago

If 9.0g of water contain 1.0 g of hydrogen what mass of oxygen is contained in 36 g of water

Mazyrski [523]

9/1 ratio so 9:1=36:x
X is your hydrogen in the 36g sample. To find the mass of Oxygen you subtract the mass of hydrogen from the total.
4 grams H
36-4=32 g h2o

4 0

3 years ago

Density is considered what type of property

Taya2010 [7]

Density is a physical property. It's measured and doesn't change the object chemically.

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3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago

What are the 5 steps of conversion​

What are the 5 steps of conversion