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3 years ago

How can global climate change be studied?

Chemistry

1 answer:

IgorC [24]3 years ago

7 0

Using physical evidence from multiple different sources, there’s lots of other ways but that’s a starting point

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You have a vial that may contain lead(II) cations or barium cations, but not both. Available to you are solutions of ammonium su

Bess [88]

Answer:

We should add, the KBr solution.

Explanation:

The vial contains Ba²⁺ and Pb²⁺

Our solutions are: (NH₄)₂ SO₄, NaF and KBr

We dissociate the solutions:

(NH₄)₂ SO₄ → 2NH₄⁺ + SO₄⁻²

NaF → Na⁺ + F⁻

KBr → K⁺ + Br-

When we add ammonium sulfate and sodium fluorine, we are adding sulfates and fluorines, but these two anions make precipitate both cations, we can not determine, if the vial contains Ba²⁺ or Pb²⁺

SO₄⁻²(aq) + Ba²⁺ (aq) → BaSO₄ (s) ↓

SO₄⁻² (aq)+ Pb²⁺(aq) → PbSO₄ (s) ↓

2F⁻ (aq) + Ba²⁺ (aq) → BaF₂ (s) ↓

2F⁻ (aq) + Pb²⁺ (aq) → PbF₂ (s) ↓

When we add the KBr, bromides only make precipitate with Pb²⁺, so the vial has only Pb²⁺ cation:

2Br⁻ (aq) + Ba²⁺ (aq) → BaBr₂ (aq)

2Br⁻ (aq) + Pb²⁺ (aq) → PbBr₂ (s) ↓

3 0

3 years ago

Nitrogen was heated from 17°C to 37°C. The original volume of nitrogen was 8.5 L. Find the new volume in liters assuming P and n

Nata [24]

Answer:

The new volume of the gas is 9.086 liters.

Explanation:

Let suppose that nitrogen has a behavior of ideal gas, the equation of state for ideal gases is:

$P\cdot V = n \cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in atmospheres.

$V$ - Volumen, measured in liters.

$n$ - Molar amount, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in atmosphere-liters per mole-Kelvin.

If pressure and molar amount of the gas remain constant, then we construct the following relationship:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$ (2)

If we know that $T_{1} = 290.15\,K$ , $P_{1} = 8.5\,L$ and $T_{2} = 310.15\,K$ , then the new volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{310.15\,K}{290.15\,K} \right)\cdot (8.5\,L)$

$V_{2} = 9.086\,L$

The new volume of the gas is 9.086 liters.

7 0

3 years ago

When heated, KClO3 decomposes into KCl and O2. KClO3--> 2KCl+ 3O . If this reaction produced 62.6 g of KCl, how much O2 was p

Vladimir [108]

The solution to your problem is as follows:

2 KClO3 → 2 KCl + 3 O2

MW of KCL = 39.1 + 35.35 = 74.6 g/mole

62.6/74.6 = 0.839 moles KCl produced

0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced

Therefore, there are 40.27g of O2 produced during the reaction.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

5 0

3 years ago

Oi oi oi oi ewin .................

KatRina [158]

Muy bein me gusta hope this makes you smile

7 0

2 years ago

The density of benzene at 15 ∘C∘C is 0.8787 g/mL. Calculate the mass of 0.1700 LL of benzene at this temperature. Express your a

SVETLANKA909090 [29]

Answer:

Mass of benzene is: 149.3 g

Explanation:

Let's use density to calculate mass.

Density = Mass / Volume

Mass = Density . volume

Be careful, because density is in g/mL and the volume is in L. So let's convert the L to mL: 0.170 L . 1000 mL / 1L = 170 mL

0.8787 g/mL . 170 mL = Mass of benzene

Mass of benzene is: 149.3 g

5 0

3 years ago