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2 years ago

why is it important for an art conservation scientist to study the properties of a painting before its repaired

1 answer:

4 0

Answer : Art conservation requires minute study for the properties of a painting before it undergoes a repair work as it may help the scientist to predict the cause of damage and the areas will be identified and accordingly the damage repair work can be started off. To maintain the overall integrity of the painting intact it needs to be studied well. For repairs to be more discrete and less obstructive it has to be inspected well so that the authenticity of the painting is not lost.

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A pupil adds 5 cm of 12 mol.dm sulphuric acids to make a 250 cmº solution. Calculate the

klemol [59]

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

Cb = ?

Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

8 0

2 years ago

Which of the following are diatomic elements? Select all that apply.

TiliK225 [7]

Answer:

Explanation:

A diatomic element in that list is Bromine

8 0

2 years ago

(Please answer as quickly as possible for brainliest, thanks :))

Eddi Din [679]

A. actinides do not occur in nature.

4 0

2 years ago

Read 2 more answers

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

3 years ago

What effect does mass have on an objects density?

xz_007 [3.2K]

The density of an object or quantity of matter is its mass divided by its volume.

6 0

2 years ago