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4 years ago

What is a Scientific law

1 answer:

6 0

A scientific law is a statement based on repeated experimental observations that describes some aspects of the universe. A scientific law always applies under the same conditions, and implies that there is a causal relationship involving its elements.

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What is the frequency of a radio wave with a wavelength of 578.0m

Radda [10]

Answer:

0.0017 Hz

Explanation:

the solution is in the photo

4 0

3 years ago

Information-

tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0

3 years ago

Un estudiante debe preparar una disolución 2M de NaCl (58,44 g/mol) en un recipiente de 7L. ?Cuántos gramos de NaCl debe agregar

Gnesinka [82]

Answer:

818.2 g.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of NaCl)/(Volume of the solution (L))

M = 2.0 M.

no. of moles of NaCl = ??? mol,

Volume of the solution = 7.0 L.

∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)

∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.

To find the mass of NaCl, we can use the relation:

no. of moles of NaCl = mass/molar mass

∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) = (14.0 mol)*(58.44 g/mol) = 818.2 g.

3 0

3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

3 years ago

PLZ HELP ME FAST I WILL GIVE BRAINLIST!! Two vehicles are traveling down the highway going 40 mph. Which has more kinetic energy

Mars2501 [29]

They actually have the same amount of kinetic energy but the bus I would say is more cuz it weighs more mass, more mass means the more energy it's going to need.

5 0

3 years ago

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