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nignag [31]
2 years ago
15

IMPORTANT IMPORTANT NEED TO GET OFF SOON BC ITS MY BROTHERS BIRTHDAY PLEASEEEE NEED THIS EXTREMELY BAD ❤️ FOR A BIG GRADE TOO

Chemistry
2 answers:
frosja888 [35]2 years ago
8 0

its d because it has to be diagnoly to the right

love history [14]2 years ago
8 0

Answer:

I think C

happy birthday to your brother though

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You are given a sample of limestone, which is mostly CaCO3, to determine the mass percentage of Ca in the rock. You dissolve the
irakobra [83]

Answer:

34.15% is the mass percentage of calcium in the limestone.

Explanation:

Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g

1 mg = 0.001 g

Moles of calcium oxalate = \frac{0.1402 g}{128 g/mol}=0.001095 mol

1 mole of calcium oxalate have 1 mole of calcium atom.

Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.

Mass of 0.001095 moles of calcium :

0.001095 mol × 40 g/mol = 0.04381 g

Mass of sample of limestone = 128.3 mg = 0.1283 g

Percentage of calcium in limestone:

\frac{0.04381 g}{0.1283 g}\times 100=34.15\%

34.15% is the mass percentage of calcium in the limestone.

7 0
2 years ago
Read 2 more answers
Is freshwater a mixture?
dem82 [27]

Answer:

No

Explanation:

A mixture is when two or more substances combine physically together. However, in water, two hydrogen atoms combine with one oxygen atom chemically, forming a new substance that has properties different from hydrogen alone or oxygen alone. ... Therefore, water is not a mixture

7 0
2 years ago
Substance A has the following properties.
ella [17]

It will take 1.11 min to heat the sample to its melting point.

Melting point = - 20°C

Boiling point = 85°C

∆H of fusion = 180 J/g

∆H of vap = 500 J/g

C(solid) = 1.0 J/g °C

C(liquid) = 2.5 J/g °C

C(gas) = 0.5 J/g °C

Mass of sample = 25 g

Initial temperature = - 40°C

Final temperature = 100°C

Rate of heating = 450 J/min

Specific heat capacity formula:- q = m ×C×∆T

Here, q = heat energy

        m = mass

        C = specific heat

      ∆T = temperature change

Melting point = - 20°C

C(solid) = 1.0 J/g °C

∆T = final temperature - initial temperature = -20 - (-40) = 20

Put these value in  Specific heat capacity formula

q = m ×C×∆T

q = 25×1.0×20

   =500J

The Rate of heating = 450 J/min

i.e. 450J = 1min

so, 500J = 1.11min

1.11 minutes does it take to heat the sample to its melting point.

The specific heat capacity is defined as the amount of heat absorbed in line with unit mass of the material whilst its temperature increases 1 °C.

Learn more about specific heat capacity here:- brainly.com/question/26866234

#SPJ4

3 0
1 year ago
Which type of intermolecular attractive force operates between (part A) all molecules, (part B) polar molecules, (part C) the hy
nydimaria [60]

Answer:

PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.

PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.

PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.

3 0
2 years ago
Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres
____ [38]

Answer:

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature =V_1=3.4 L

Number of moles in the container = n_1

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = V_2=5.11 L

Number of moles in the container after addition=n_2=2.28\times 10^{-2} mol

Applying Avogadro's law:

\frac{Volume}{Moles}=constant (at constant pressure and temperature)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}

n_1=1.51\times 10^{-2} mol

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

8 0
3 years ago
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