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MaRussiya [10]
3 years ago
14

3,560 into scientific notation

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

3.56 x 10^3

Explanation:

In order to put in in scientific notation, the number must be between 1-10 and must be multiplied by a power of 10. We can turn 3560 into a number 1-10 by moving the decimal up.

We move it up 3 spaces. 3.560

Since we moved it up 3 spaces, we need to multiply it by 10^3

3.56 x 10^3

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Using the table below, what is the change in enthalpy for the following reaction? 3CO (g) + 2Fe2O3 (s) Imported Asset Fe(s) + 3C
zhuklara [117]

To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

ΔH = Total H of products – Total H of reactants

You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

ΔHf CO2 (g) = -393.5 kJ/mol

ΔHf CO (g) = -110.5 kJ/mol

ΔHf Fe2O3 (s) = -822.1 kJ/mol

ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0 kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>

ΔH = <span>795.2 kJ</span>

3 0
2 years ago
You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,20}, thus,

Thus, you need to add:

[H⁺] = 10^{-7,2} -10^{-8,0} = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × \frac{1H_{2}SO_4 mol}{2H^{+} mol} = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × \frac{98,1 g}{1 mol} × \frac{100 gSolution}{12 gH_{2}SO_4 } × \frac{1 mL}{1,080 g}  =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0
3 years ago
The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.
Alona [7]

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

3 0
3 years ago
Why might nuclear fusion be a better source of energy on earth than nuclear fission?
AleksAgata [21]
<span>Cost and availability of fuel is a considerable factor when dealing with nuclear power. Fission requires an element that can be easily split in a particle accelerator, such as uranium or plutonium. Fusion, on the other hand, uses isotopes of hydrogen atoms, specifically deuterium and tritium, that can be obtained from ordinary water</span>
5 0
3 years ago
If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

5 0
3 years ago
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