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Sliva [168]
2 years ago
11

I NEED HELP NOW!!!!!!!!!!!

Chemistry
1 answer:
Rashid [163]2 years ago
3 0

A: the ball in frame A had the highest velocity, and the ball in frame B has the highest kinetic energy

You might be interested in
A sample of TNT, C7H5N3O6 , has 8.94 × 1021 nitrogen atoms. How many hydrogen atoms are there in this sample of TNT?
Bess [88]

Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

8 0
3 years ago
Density is d= m/vol. If a material has a mass of 65.5 g and a volume of 32.5 ml, it has a density of
ryzh [129]

Using the given formula, the density of the material is 2.015 g/mL

<h3>Calculating Density </h3>

From the question, we are to determine the density of the material

From the given formula

Density = Mass / Volume

And from the given information,

Mass = 65.5 g

and volume = 32.5 mL

Putting the parameters into the equation,

Density = 65.5/32.5

Density = 2.015 g/mL

Hence, the density of the material is 2.015 g/mL.

Learn more on Calculating density here: brainly.com/question/24772401

#SPJ1

8 0
2 years ago
The atomic mass of Cu is 63.5. Find its electrochemical equivalent​
FrozenT [24]

Answer:

The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

m = Z·I·t = Z·Q

We have;

Cu²⁺ (aq) + 2·e⁻ → Cu

Therefore, one mole of Cu, is deposited by 2 moles of electrons

The charge carried one mole of electrons = 1 Faraday = 96500 C

∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

∴ Z = 3.29015544 × 10⁻⁴ g/C = 3.29015544 × 10⁻⁷ g/C

The electrochemical equivalent of copper, Cu, is Z = 3.29015544 × 10⁻⁷ g/C

7 0
3 years ago
Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
2 years ago
A mixture of 117.9 g of P and 121.8 g of O, reacts completely to form P4O6 and P4O10. Find the masses of P4O6 and P4O10
Alex17521 [72]

Answer:

Mass of P4O6=103.4

            P4O10=133.48

Explanation:

Balanced reaction is:

8P +8O_{2}  ⇒P_{4} O_{6} +P_{4} O_{10}

Both reactant completely vanishes as equivalent of bot are equal.

Moles of P=\frac{117.9}{31} =3.80

Moles of O_{2} =\frac{117.9}{31} =3.80

No. of moles of formed product are equal and is \frac{1}{8}th of mole of any of reactant.

Thus weight of  P_{4} O_{6} =\frac{3.80}{8}×220 ≈103.41

        weight of  P_{4} O_{10} =\frac{3.80}{8}×284 ≈133.48

4 0
2 years ago
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