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3 years ago

Compute 4.659×104−2.14×104. Round the answer appropriately.

2 answers:

7 0

<span>To compute 4.659×104−2.14×104, the first step is the factorization. That is as follows:4.659×104−2.14×104= 10^4.(4.659−2.14), the next step is to compute 4.659−2.14=2.51, so 10^4.(4.659−2.14)=2.51x10^4=2.51x 10000 (because10^4=10000), the last calculus is 2.51x 10000=25100, the final answer is 25,000.Hope this helps. Let me know if you need additional help!</span>

Effectus [21]3 years ago

3 0

Answer is: proper number of significant figures is 25190.00.

10⁴ = 10×10×10×10.

10⁴ = 10000.

1) 4.659×10⁴ = 4.659×10000.

4.659×10⁴ = 46590.

2) 2.14×10⁴ = 2.14×10000.

2.14×10⁴ = 21400.

3) 4.659×10⁴ - 2.14×10⁴ = 46590 - 21400.

4.659×10⁴ - 2.14×10⁴ = 25190 (2.519×10⁴).

An integer is a number that can be written without a fractional component.

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A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago

What is the identity of the atom shown?

jonny [76]

Niterogen if I’m wrong please correct me

3 0

2 years ago

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(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$

Therefore, for this problem the step by step explanation is:

$P_{1} xV_{1} = P_{2} xV_{2}$

Clearing P2 and replacing

$P_{2}= \frac{P_{1} xV_{1}}{V_{2} } = \frac{4atmx3L}{20L} = 0.6atm$

7 0

2 years ago

Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

7 0

3 years ago

The gram formula mass of NH4Cl is(1) 22.4 g/mole (3) 53.5 g/mole(2) 28.0 g/mole (4) 95.5 g/mole

g100num [7]

The answer is (3) 53.5 g/mol. The gram formula mass means that the mass of one mol compound. 1N=14, 4H=4, 1Cl=35.5. So the gram formula mass of NH4Cl=14+4+35.5=53.5 g/mol.

7 0

3 years ago

Read 2 more answers