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2 years ago

How does air bubbles affect density of a solid object?

1 answer:

6 0

it decreases the density of the object the air bubbles take up space. it increases the volume of the object slightly but the objects weight remains the same, hence the objects density decreases

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In addition to DNA, what else is

Pepsi [2]

Answer: ribosomes and membranes.

Explanation:

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2 years ago

A chemist wants to calculate the amount of heat that is absorbed by a sample of copper as it is melted. Which constant should sh

34kurt

The constant used for the absorption of heat by the sample in melting is $+\Delta H_{fus}$ . Thus, option A is correct.

The chemical reaction has been defined as the energy in which the energy has been released or absorbed for the breaking of bonds in the reactants and the formation of product.

<h3>Constant for energy absorbed</h3>

The energy has been absorbed in the melting of the copper sample. Thus, the sample has been converted from the solid to the liquid state.

The change in energy with the conversion in solid and liquid state has been termed as heat of fusion.

The energy has been absorbed by the system, thus it has been marked with the positive sign.

Therefore, $+\Delta H_{fus}$ has been the constant used for the absorption of heat by the sample in melting. Thus, option A is correct.

Learn more about melting sample, here:

brainly.com/question/8828503

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2 years ago

Which has the higher percentage of aluminum, Al2O, or Al(NO3)3

Rzqust [24]

Al₂O has a higher percentage of aluminium.

Explanation:

To solve this problem, we have to compare the molar mass of the aluminium in each compound to one another as percentage of the whole compound:

Molar mass of Al₂O = 2(27) + 16 = 70g/mol

Molar mass of Al(NO₃)₃ = 27 + 3[14 + 3(16)] = 213g/mol

Percentage by mass of Al in Al₂O = $\frac{2x27}{70}$ x 100 = 77%

Percentage by mass of Al in Al(NO₃)₃ = $\frac{27}{213}$ x 100 = 12.7%

Al₂O has a higher percentage of aluminium.

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

8 0

3 years ago

A stack of 10 pennies weighing 26.55 grams is placed in a graduated cylinder containing water. The initial volume of water is 50

BigorU [14]

In order to determine the density of an item, we will need to determine its mass and volume. The standard unit for measuring mass in a lab is the gram. Think about liquids- what units do you typically report the volume of a liquid in? What about for a sugar cube, what volume is the most appropriate?
A regular object like a sugar cube can be measured with a ruler so we might report the volume in centimeters cubed (cm3). An irregular object like the plate pictured below can be measured by using a technique called volume by displacement. A liquid (typically water) is placed in a graduated cylinder and the volume of a liquid is measured. Then the irregular object is placed in the liquid and the volume is measured again. The change in volume is the irregular object’s volume. This measurement is often made using a graduated cylinder and recording a volume in Liters or milliliters (mL).
Figure 1. (a) Regular object of metal blocks with the same width, length, and height. (B) An irregular

7 0

3 years ago

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