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irina1246 [14]
2 years ago
13

Explain why science is is you significant digits

Chemistry
1 answer:
-BARSIC- [3]2 years ago
7 0
Why use significant digits you ask? Significant digits or significant figures are use as to not be too precise of an actual calculated measure.
Long decimal numbers location like 3.45 * sqrt of 3 gives 5.975575286112627.
That number is too precise.

Rounding it to 5.98 shows an exact closeness to the precision of the measure.
You might be interested in
Determine the percentage of carbon and hydrogen in ethane C2H6 if the molecular weight is 30.
Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
PLEASE HELP!! GIVING BRAINLIEST
RSB [31]

Answer:

The answer is "0.00172172603".

Explanation:

Given:

Mass (M) = 1.60 \times  10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) =  ?

Formula:

\to \bold{V = \frac{M}{D}}

       = \frac{1.60 \times  10^{-3}}{9.293 \times  10^{-1}} \\\\  = \frac{1.60 \times  10^{1}}{9.293 \times  10^{3}} \\\\ = \frac{1.60 \times  10}{9.293 \times  1000} \\\\ = \frac{1.60 }{9.293 \times  100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603

4 0
2 years ago
A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
Lana71 [14]

Answer:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol

Moles of hydrogen gas produced = 0.01225 mol

2M+2xHCl\rightarrow 2MCl_x+xH_2

Moles of metal =\frac{0.225 g}{27.0 g/mol}=8.3333 mol

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

2M+6HCl\rightarrow 2MCl_3+3H_2

MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

3 0
3 years ago
13. Sodium, Na, shows metallic properties. Identify another element that is likely to show similar metallic properties.
Deffense [45]

Answer:

19.K, potassium

Explanation:

it has all properties of metals

7 0
2 years ago
A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

6 0
3 years ago
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