Calculate the density of softdrinks in g/cm3.240 g Weight of an empty beaker420 g Weight of an empty beaker with softdrinks. g Weight of softdrinks230 ml Volume of softdrinks_g/cm3 Density​

1. Calculate the density of softdrinks in g/cm3.240 g Weight of an empty beaker420 g Weight of an empty beaker with softdrinks. g Weight of softdrinks230 ml Volume of softdrinks_g/cm3 Density​

The density of soft drinks : ρ = 0.783 g/cm³

Further explanation  

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

The unit of density can be expressed in g/cm³ or kg/m³  

Density formula:  

[tex]\tt \rho=\dfrac{m}{V}[/tex]

ρ = density  

m = mass  

V = volume  

Given

240 g Weight of an empty beaker

420 g Weight of an empty beaker with soft drinks

230 ml(230 cm³) Volume of soft drinks

Required

The density of soft drinks

Solution

mass of soft drinks :

= 420 g - 240 g

= 180 g

Input the value in the formula :

ρ = 180 g : 230 cm³

ρ = 0.783 g/cm³

Learn more

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2. 12. What is the mass of 300 mL of benzene if its density of benzene is 0.88 g/mL? A. 264 g B. 300.88g C. 299.12 g D. 340.90 g 13. Three identical pieces of stone were dropped in graduated cylinder with 50 mL of water. Afterwhich, the volume of water rose to 96.5 mL. What is the volume of EACH stone? A. 146.5 mL B. 15.5 mL C. 32.17 mL D. 52.5 mL 14. What volume of silver metal will have a mass of exactly 2500 grams? The density of silver is 10.5 g/cm3. A. 238.1 cm3 B. 2489.5 cm3 C. 2510.5 cm3 D. 26,250 cm3 PLease po need ko na po now na po

Answer:

12c

13a

14c

Explanation:

Sana maka tulong

3. Ethanol has a density of 0.789 g/cm3 . what is the mass of a 25.0 ml sample of ethanol

Answer:

D kayo Alana said I don't

Explanation:

Know what I want to go back and I have done it before.

4. The density of gold is 19.3 g/cm3. What is the mass of 11.3 cm3 of gold?

Answer:

218.09g

Explanation:

5. The density of platinum is 23.4 g/cm3 . Calculate the mass of 75.0 cm3 of platinum.

Answer:

Hope you understand my solution. I'm a newbie here!

Explanation:

Density = Mass / Volume

Mass = Density x Volume

Mass = 23.4 g/cm3 x 75.0cm3

Mass = 1755 grams

6. A container has a volume of 40 mL and a density of 0.89 g/cm3. What is the mass of the container?

[tex]\tt{\huge{\blue{Explanation:}}}[/tex]

The density of an object is given by

[tex]\boxed{d = \frac{m}{V}}[/tex]

where:

d = density

m = mass

V = volume

[tex]\tt{\huge{\red{Solution:}}}[/tex]

Solving for m

Note: 40 mL = 40 cm³

[tex]d = \dfrac{m}{V}[/tex]

m = dV

m = (0.89 g/cm³)(40 cm³)

[tex]\boxed{m = \text{35.6 g}}[/tex]

Therefore, the mass of the container is 35.6 g.

[tex]\\[/tex]

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7. 8. The Earth's crust has a density ofand a thickness ranging froma. 2.8 g/cm3: 5-50 kmc. 3.0 g/cm3; 5 -60 kmb. 2.9 g/cm3: 5-50 kmd. 3.1 g/cm3; 5-60 km​

Answer:

a.2.8 g/cm3: 5-50 km

Explanation:

because the earth crust has 6.0g/ that will devide to the earth mantle2.3 its equal to letter ( a.)

8. What is the volume of 11.3 g graphite, density of 2.25 g/cm3.​

Answer:

11.3 × 2.25 = 25.425 g/cm3

Explanation:

PA BRAINLIEST AT PA FOLLOW

9. You have a sample of granite with density 2.8 g/cm3. The density of water is 1.0 g/cm3. What is the specific gravity of your granite? ​

Answer:

2.8

Explanation:

10. THE MASS OF AN OBJECT IS 26 GRAMS. CALCULATE ITS DENSITY IF IT HAS A VOLUME OF 16 CM3? 1. 1.52 G\CM3 2. 1.62 G\CM3 3. 16 G\CM 4. NONE OF THE A BOVE

Answer:

None of the above

Explanation:

Pa brainliest din po ty <3

11. How many atoms, molecules, or formula units of the substance are in the following amounts? a) 3.50 mol of O2 b) 2.75 g of S8 c) 5.50 g of C12H22O11 d) 5.00 g of Al(NO3)3 e) 5.00 mL of ethanol, CH3CH2OH, density = 0.790 g/mL f) a sphere of chromium metal, 0.343 mm in diameter. The density of chromium is 7.20 g/cm3​

Answer:

The density of solid crystalline chromium is 7.18 g/cm3.

Explanation:

Correct me if im wrong

12. What is the volume of 22 g of gold, which has a density of 19.3 g/cm3?

Chemistry: Volume of Gold

To find the volume of a 22 grams of gold, divide the mass into its density.

Given:

[tex] \sf mass \: of \: gold \: = \: 22 \: grams[/tex]

[tex] \sf density \: of \: gold \: = \: 19.3 \: g/cm {}^{3} [/tex]Substitute:

[tex] \sf v = \frac{mass}{density} [/tex]

[tex] \sf v = \frac{22 \: grams}{19.3 \: g/ {cm}^{3} } [/tex]

[tex] \sf v \approx 1.13 \: cm {}^{3} [/tex]

Hence, the volume of the gold is approximately 1.13 cm³.

[tex] \green ☘ [/tex]

13. cm3 of oil weighs 0.9 g. What is the weight of 20 cm3 of oil?​

ANSWER:

18g

hope it helps! :)

14. what is the mass of a ring if it's density is 2.5g/cm3 and it's volume is 0.5 cm3 ? choices: a. 1.8 g b. 1.25 g c. 1.4 g d. 1.14 g

Answer:

letter b.

Explanation:

The density of an object is given by

d m V

where:

d = density

m = mass

V = volume

Solution:

Solving for m

d m V =

m = dV

m = (2.5 g/cm³)(0.5 cm³)

m = 1.25g

Therefore, the answer is b. 1.25 g

15. 14. If the blackboard eraser has a mass of 180 grams and the volume is 144 cm", itsdensity isa. 2.25 g/cm3C. 1.25 g/cm3b. 3.25g/cm3d. 1.40 g/cm3​

Answer:

C

Explanation:

If you divide mass and volume you will get the answer

16. 8. 6.25 mL = ______dL 9. 276 cm3 = _______ m3 10. 75 kg = _______ g

Answer:

0.0625 decilitre0.000276 cubic metre75000 grams

17. Cubes are three-dimensional square shapes that have equal sides. What is the density of a cube that has a mass of 12.6 g and a measured side length of 4.1 cm? (Density: D = ) .1828 g/cm3 .3254 g/cm3 3.073 g/cm3 68.92 g/cm3

Answer:

It's 1828 g/cm3

Explanation:

Hope it helps!

Answer:

it's .1828 g/cm3

Explanation:

the formula for how to find the volume for a cube is V = a³ and a is where you input the side length, which I got approximately 68.92.  Once you get the volume, you do mass/volume = density, which is 12.6/68.92. Then you get your answer! which is .1828 g/cm³!

18. The density of platinum is 23.4 g/cm3. What is the volume of 105.3 g of this element?

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & m = \text{105.3 g} \\ & d = \text{23.4 g/cm}^3 \end{aligned}[/tex]

Step 2: Calculate the volume.

[tex]\begin{aligned} V & = \frac{m}{d} \\ & = \frac{\text{105.3 g}}{\text{23.4 g/cm}^3} \\ & = \boxed{\text{4.50 cm}^3} \end{aligned}[/tex]

Hence, the volume of 105.3 g of platinum is 4.50 cm³.

[tex]\\[/tex]

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19. Bulk density = 1.31 g/cm3and particle density = 2.64 g/cm3.​

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fsfsf

Step-by-step explanation:

20. A certain metal alloy is created by mixing aluminum and iron. The density of aluminum is 2.80 g/cm3 and the density of iron is 7.86 g/cm3. The density of the alloy is 5.0 g/cm3. What mass in grams of aluminum is in 30.0 cm3 of the metal alloy? Give your answer to 2 significant digits.

Answer:

To solve this problem, you can use the concept of density, which is the mass of an object per unit volume. Specifically, you can use the formula:

density = mass / volume

You know the density of the alloy and the volume of the alloy, and you want to find the mass of aluminum in the alloy. So, you can rearrange the formula to:

mass = density x volume

Therefore, you can use this formula to find the mass of aluminum in 30.0 cm3 of the alloy:

mass = 5.0 g/cm3 x 30.0 cm3 = 150 g

So, there is 150 grams of aluminum in 30.0 cm3 of the metal alloy.

Explanation:

Pa brainliest

21. 14. A student needs to find the density of a cube. Each sideof the cube measures 3cm and the mass of the cube is 12g.What is the approximate density of the cube?A. 0.4 g/cm3C. 2.25 g/cm3B.4.0 g/cm3D.36 g/cm3​

Answer:

= mass/volume

[tex] = \frac{12 \: g}{ {(3 \: cm)}^{3} } [/tex]

[tex] = \frac{12 \: g}{27 \: {cm}^{3} } [/tex]

=0.44 g/cm3

Answer is A

22. A block of wood has density 0.50 g/cm3 and mass 1 500 g. It floats in a container of oil (the oil's density is 0.75 g/cm3). What volume of oil does the wood displace? Select one: a.3 000 cm3 b.2 000 cm3 c.1 125 cm3 d.750 cm3 e.375 cm3

Explanation:

hope it helps pa brainles po please

23. The density of gold is 19.3 g/cm3. What is the volume of a 13 g gold nugget? (Density: D = ) .25 cm3 .67 cm3 1.48 cm3 2.50 cm3

Answer:

. 67 cm3

Explanation:

Given:

Density of gold = 19.3 g/cm3

Mass of gold = 13 g

To calculate:

Volume of gold

Explanation:

Density of a given substance expressed as the mass of the substance present in a unit volume

Density = mass/volume

volume = mass/density

            = 13 g/ 19.3 g.cm-3 = 0.674 cm-3

#Answerfortrees

24. 4. The density of Gold is 19.3 g/cm3. What is the volume of a 96.5 g Gold? A. 0.2 cm3 B. 5 cm3 C. 1,862 cm3 D. 1,862.45 cm3

Answer:

5 cm3

If the 96.5 grams of gold has a volume of 5 cm3

so letter b

Answer:

D ang sagot

Explanation:

yan sinabi ni kuya eh

25. what type of measurement is indicated by each of the following units? choices are in the last column 1 .g/ml 2. s 3. km 4. g 5.cm3 6. mm 7. mg 8. L 9. g/cm3 choices: density length mass time volume

Answer:

g/ml - density

g/ml - densitys - time

g/ml - densitys - timekm - length

g/ml - densitys - timekm - lengthg - mass

g/ml - densitys - timekm - lengthg - masscm3 - volume

g/ml - densitys - timekm - lengthg - masscm3 - volumemm - length

g/ml - densitys - timekm - lengthg - masscm3 - volumemm - lengthmg - mass

g/ml - densitys - timekm - lengthg - masscm3 - volumemm - lengthmg - massL - volume

g/ml - densitys - timekm - lengthg - masscm3 - volumemm - lengthmg - massL - volumeg/cm3 - density

Answer:

1. g/mL - density

2. s - time

3. km - length

4. g - mass

5. cm³ - volume

6. mm - length

7. mg - mass

8. L - volume

9. g/cm³ - density

26. given: paper size : 25" x 38" BOPP = 0.903 g/cm3 LDPE = 0.91 g/cm3 What is the thickness of 12gsm, 15gsm application of LDPE?

Answer:

To find the thickness of LDPE for 12gsm and 15gsm applications, we need to use the formula:

Thickness = (Grams per square meter / (Density x Area)) * 10000

where:

Grams per square meter is the weight of the material per unit area (in gsm)

Density is the density of the material (in g/cm3)

Area is the area of the material (in cm2)

10000 is a conversion factor to convert cm2 to m2

Given:

Paper size: 25" x 38"

BOPP density: 0.903 g/cm3

LDPE density: 0.91 g/cm3

To calculate the area of the paper, we need to convert the size from inches to centimeters:

25 inches = 25 x 2.54 = 63.5 cm

38 inches = 38 x 2.54 = 96.52 cm

Area = length x width = 63.5 cm x 96.52 cm = 6134.62 cm2

For 12gsm LDPE application:

Thickness = (12 / (0.91 x 6134.62)) * 10000

Thickness = 0.00190 cm

For 15gsm LDPE application:

Thickness = (15 / (0.91 x 6134.62)) * 10000

Thickness = 0.00238 cm

Therefore, the thickness of 12gsm LDPE application is 0.00190 cm and the thickness of 15gsm LDPE application is 0.00238 cm.

27. 7,01875 g/cm3 change into kg/m3 .

7,01875 g/cm*3 ---> 7,01875 x 10*6 / 10*3=7018,75kg/m*3

28. What is the volume of 11.3 g graphite, density of 2.25 g/cm3.​

Answer:

The volume is 5.02cm³.

Explanation:

The formula of density is: density (d) = mass (m)/ volume (v)

So, to calculate the volume from density and mass we use the density formula by making a few math steps:

From this: d = m/v

- The volume pass to the other side of the formula as the density goes dividing the mass: v= m/d

- Then, replace the information: v = 11.3g/

(2.25g/cm³)

- By dividing, we find that the volume is v = 5.02cm³

29. A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate the mass percentage, mole fraction, molality, and molarity of H2SO4 in the solution. Given: Mass H2SO4 = 571.6 g Volume of Solution = 1 L Density of solution = 1.329 g/cm3 = 1.329 g/mL Required: a)Mass % b)Mole fraction c)Molality d)Molarity

Solution (a)

Step 1: Calculate the mass of solution.

[tex]\text{mass of solution = density of solution × volume of solution}[/tex]

[tex]\text{mass of solution = 1.329 g/m L × 1000 m L}[/tex]

[tex]\text{mass of solution = 1329 g}[/tex]

Step 2: Calculate the mass percent.

[tex]\text{mass percent} = \frac{\text{mass}_{H₂SO₄}}{\text{mass of solution}} × 100[/tex]

[tex]\text{mass percent} = \frac{\text{571.6 g}}{\text{1329 g}} × 100[/tex]

[tex]\boxed{\text{mass percent = 43.01\%}}[/tex]

Solution (b)

Step 1: Calculate the mass of water.

[tex]\text{mass}_{water} = \text{mass of solution} - \text{mass}_{H₂SO₄}[/tex]

[tex]\text{mass}_{water} = \text{1329 g} - \text{571.6 g}[/tex]

[tex]\text{mass}_{water} = \text{757.4 g}[/tex]

Step 2: Calculate the molar mass of H₂SO₄ and water.

For H₂SO₄

molar mass = (1.008 g/mol × 2) + (32.07 g/mol × 1) + (16.00 g/mol × 4)

molar mass = 98.09 g/mol

For water

molar mass = (1.008 g/mol × 2) + (16.00 g/mol × 1)

molar mass = 18.02 g/mol

Step 3: Calculate the number of moles of H₂SO₄ and water.

For H₂SO₄

[tex]n_{H₂SO₄} = \text{571.6 g H₂SO₄} × \frac{\text{1 mol H₂SO₄}}{\text{98.09 g H₂SO₄}}[/tex]

[tex]n_{H₂SO₄} = \text{5.827 mol}[/tex]

For water

[tex]n_{water} = \text{757.4 g water} × \frac{\text{1 mol water}}{\text{18.02 g water}}[/tex]

[tex]n_{water} = \text{42.03 mol}[/tex]

Step 4: Calculate the mole fraction of H₂SO₄.

[tex]X_{H₂SO₄} = \frac{n_{H₂SO₄}}{n_{H₂SO₄} \: + \: n_{water}}[/tex]

[tex]X_{H₂SO₄} = \frac{\text{5.827 mol}}{\text{5.827 mol + 42.03 mol}}[/tex]

[tex]\boxed{X_{H₂SO₄} = 0.1218}[/tex]

Solution (c)

[tex]\text{molality} = \frac{n_{H₂SO₄}}{\text{mass}_{water} \: \text{(kg)}}[/tex]

[tex]\text{molality} = \frac{\text{5.827 mol}}{\text{0.7574 kg}}[/tex]

[tex]\boxed{\text{molality} = \text{7.693 m}}[/tex]

Solution (d)

[tex]\text{molarity} = \frac{n_{H₂SO₄}}{\text{volume of solution (L)}}[/tex]

[tex]\text{molarity} = \frac{\text{5.827 mol}}{\text{1 L}}[/tex]

[tex]\boxed{\text{molarity} = \text{5.827 M}}[/tex]

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30. What is 5 g/L to kg/cm3​

Answer:

5000 - 0.005

divide the mass/volume value by 1e+6