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3 years ago

Is 6.0 × 10 to the negative 4th power is a significant figure?

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

Yes

Explanation:

First remember that a significant figure are basically values that contribute to the precision of a value. In any scientific notation the values are significant figures because these values stay the same. In this case we have two significant figures which is 6 and 0, you can further prove that they are significant figures by converting the notation into standard form.

$6.0\times10^{-4}$

Negative so move the decimal point to the left:

$=0.00060$

6 and 0 are the significant figures in this standard notation because it's precise to it's actual value which is 6.0.

To sum up, the values you have on the left side of a notation are significant figures since they will not change no matter if it's standard or scientific notation meaning it's precise.

Hope this helps.

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How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in theelectrolysis of molten MgCl2 with an appli

Llana [10]

Answer:

There is 29.8 kilowatt hours needed.

Explanation:

Step 1: Data given

Mass of Magnesium = 3.00 kg

Molar mass of magnesium = 24.31 g/mol

Applied emf = 4.50 V (= 4.50 J/C)

Step 2: Calculate moles of Magnesium

Moles = Mass Mg / Molar mass Mg

Moles Mg = 3000 grams / 24.31 g/mol

Moles Mg = 123.4 moles

Step 3: Calculate how many electrons are needed to produce the magnesium.

The ionic equation for the reduction of Mg^2+ :

Mg^2+ + 2e^- → Mg

Every mole of Mg requires 2 mol of electrons.

For 123.4 mol of Mg, we have 246.8 mol of electrons.

Step 3: Find how many coulombs are involved.

The Faraday constant = 96500 couloumbs

1 mole of electrons is 96500 coulombs.

246.8 mol of electrons need 2.38 *10^7 Coulombs

Step 4: Calculate kilowatt-hours of electricity needed

2.38 * 10^7 C * 4.5 J/C = 10.7 * 10^7 J

10.7 * 10^7 J * ( 1 kW-h-/ 3.6*10^6 J ) = 29.8 kWh

There is 29.8 kilowatt hours needed.

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3 years ago

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3 years ago

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The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

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3 years ago

A crime lab received a 235-gram sample. The sample had a molecular mass of 128.1 grams and the empirical formula is CH2O. How ma

kvv77 [185]

Explanation:

Given parameters:

Mass of sample = 235g

Molecular mass of sample = 128.1g

Empirical formula = CH₂O

Unknown:

Mass of each element in the sample = ?

Solution:

To solve this problem, we must know that the empirical formula of any compound is the simplest ratio of the atoms it contains. This is not the true formula of the compound.

Molecular formula = (Empirical formula)ₙ

Let us find the molecular mass of the sample;

CH₂O = 12 + 2(1) + 16 = 30g

128.1 = (30)n

n = 4

The molecular formula of the compound is; (CH₂O)₄ = C₄H₈O₄

Now to find the grams of each element in the sample;

Express the molecular mass of each element and that of the compound as a fraction and multiply with the given mass;

For C;

$\frac{4 x 12}{128.1} x 235\\$ = 88.06g