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bulgar [2K]
2 years ago
5

Read the examples below and determine if it describes speed, velocity or acceleration.

Chemistry
1 answer:
Zarrin [17]2 years ago
8 0

Answer:

hmmmmmmmmmmmmmmmmmmm

Explanation:

not sure tbh, ask ur teacher unless someone else helps you!

You might be interested in
I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19
IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

For the 4.12 g sample

Moles of a substance is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}

Number of molecules is given by

nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}

For the 19.37 g sample

n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}

Number of molecules is given by

nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}

1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0
3 years ago
What is insulin's structure?<br> a.quaternary<br> b.secondary<br> c.primary<br> d.tertiary
svetoff [14.1K]

Answer:

A

Explanation:

Quaternary structure of proteins is composed of two or more polypeptide chains. Insulin has two; one alpha  and one beta chain. The two chains are joined together by disulfide bonds at two points (at cysteines). Other examples of quaternary proteins structures are DNA polymerase and hemoglobin.

7 0
2 years ago
Name any <br>three kinds of mixtures with example​
sleet_krkn [62]

Answer:

Here are a few more examples:

Smoke and fog (Smog)

Dirt and water (Mud)

Sand, water and gravel (Cement)

Water and salt (Sea water)

Potassium nitrate, sulfur, and carbon (Gunpowder)

Oxygen and water (Sea foam)

Petroleum, hydrocarbons, and fuel additives (Gasoline)

Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.

Examples of Heterogeneous mixtures – air, oil, and water, etc.

Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.

Explanation:

8 0
2 years ago
Read 2 more answers
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
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