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marshall27 [118]
2 years ago
11

Help i attached question

Chemistry
2 answers:
Oksanka [162]2 years ago
6 0

Answer:

bat dipo maklaro ang pic po

Inga [223]2 years ago
3 0

Answer:

option D is correct

Explanation:

hope it helps you

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Which law is based on the graph that is shown below?
Cerrena [4.2K]

Answer:

Boyle's Law

Explanation:

8 0
3 years ago
Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc
Ber [7]

Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

  • 1 mole of CH4 → 4 mole of 4 mole of fluorine
  • 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

  • 4 moles of  fluorine →  1 mole of CF4
  • 0.56 mole →  \frac{1}{4} * 0.56 = 0.14mole of CF4
  • 4 moles of fluorine →  4 moles of HF
  • 0.56 mole of fluorine →  0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0
3 years ago
PLEASE HELP ME ASAP! PLEASE PLEASE PLEASE!
dsp73
It’s C because the oak trees create a population
4 0
3 years ago
Read 2 more answers
A standard lanthanum solution is prepared by dissolving 0.1968 grams of lanthanum oxide (La2O3) in excess nitric acid and diluti
sweet-ann [11.9K]

Answer:

1.208x10⁻³M and 392.5ppm La(NO3)3

Explanation:

The reaction that occurs is:

La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O

Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.

To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:

<em>Moles La2O3 -Molar mass: 325.81g/mol-</em>

0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3

<em>Moles La(NO3)3:</em>

6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3

<em>Molarity:</em>

1.208x10⁻³ moles La(NO3)3 / 1L =

<h3>1.208x10⁻³M</h3>

<em>Mass La(NO3)3 -Molar mass: 324.92g/mol-</em>

1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3

In mg:

392.5mg La(NO3)3 / 1L =

392.5ppm La(NO3)3

7 0
2 years ago
How much heat is required to raise 36 g ice at – 10.0oC to steam at 110oC? (get your answers from question #1)
Gnom [1K]

Answer:

The total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

Explanation:

The heat involved in this process involves the following:

1. Heat to change ice at -10°C to ice at 0°C;

2. Heat to change ice at 0°C to water at 0°C

3. Heat to change water at 0°C to water at 100°C

4. Heat to change water at 100°C to steam at 100°C

5. Heat to change steam at 100°C to steam at 110°C

Specific heat capacity of ice, c = 2040 J/K/kg, Latent heat of fusion of ice, L = 3.35 × 10⁵ J/kg, specific heat capacity of water, c =  4182 J/K/kg, latent heat of vaporization of water, l = 2.26 × 10⁶ J/kg, specific heat capacity of steam, c = 1996 J/K/kg

Step 1: H = mcθ; where m = 30.0 g = 0.03 g, c = 2040 J/K/kg, θ = (0 -  -10) = 10 K

H = 0.03 * 2040 * 10 = 612 J

Step 2: H = mL, where  L = 3.35 × 10⁵ J/kg

H = 0.03 * 3.35 × 10⁵ = 10050 J

Step 3: H = mcθ, where c =  4182 J/K/kg, θ = (100 - 0) = 100 K

H = 0.03 * 4182 * 100 = 12546 J

Step 4: H = ml, where l = 2.26 × 10⁶

H = 0.03 * 2.26 × 10⁶ = 67800 J

Step 5: H = mcθ, where c = 1996 J/K/kg, θ = (110 - 100) = 10 K

H = 0.03 * 1996 * 10 = 598.8 J

Total heat required to raise ice at -10°C to steam at 110°C = (612 + 10050 + 12546 + 67800 + 598.8) J = 91606.8 J

Therefore, the total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

8 0
2 years ago
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