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Llana [10]
3 years ago
15

There are several hazards associated with the chemicals in the qualitative analysis experiment. for each chemical, identify the

associated hazards. (select all that apply.) hazard 0.1 m ag+ 0.1 m ba2+ 0.1 m fe3+ 6 m hcl 6 m h2so4 6 m hno3 7.5 m nh3 corrosive irritating vapors toxic if ingested causes staining on skin flammable none listed
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0
You can determine the hazards of these chemicals by looking at their material data safety sheets (MSDS). 

1. 0.1 M Ag⁺: Silver compounds are absorbed by skin causing bluish pigmentation. Thus, it <em>causes </em><span><em>staining on skin</em>.</span> Also, liquid <em>vapor may be irritating</em> to skin and also <em>moderately toxic when ingested</em>.

2. 0.1 M Ba²⁺: This is <em>mildly toxic when ingested</em> causing stomach irritation, muscle weakness, swelling of organs like brain, liver, kidney and heart.

3. 0.1 M Fe³⁺:Iron is <em>corrosive, has irritating vapor especially to the eyes, and toxic if ingested</em>.

4. 6 M HCl: This is a concentrated strong acid, so it is <em>corrosive, has irritating vapors, flammable and toxic when ingested</em>.

5. 6 M H₂SO₄: This is also a concentrated strong acid. Moreover, it is a strong oxidizing agent. So, its hazards include: <span><em>corrosive, has irritating vapors, toxic when ingested and causes staining on skin</em>.

</span>
6. 6 M HNO₃: This is a concentrated strong acid, so it is <em>corrosive, has irritating vapors, flammable and toxic when ingested</em>.

7. 7.5 M NH₃: This is a weak base. It is characterized for its pungent odor. This is <em>corrosive, has irritating vapors, toxic if ingested, and flammable</em>.
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Answer:

D- a weak base

Explanation:

The ph scale goes from 1-10 a solution over 7ph is classified as basic. A solution thats 8.4 is only 1.4 over 7pH, making it a weak basic solution. An example of a strong base would be a solution with a pH of 9.2 (for example).

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3 years ago
Which choice is not a property of an acid?
Lelu [443]

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it's going to be D

Explanation:

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Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of
Alenkinab [10]

Answer:

Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together.  After the big bang,<em> the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently</em>.

The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.

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3 years ago
A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe
nikitadnepr [17]

Answer:

\large \boxed{34.2\, ^{\circ}\text{C}}

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:  

For the metal:

m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

For the water:

m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}

\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}

\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

3 0
3 years ago
What is the mass, in grams, of a sample of 6.98 × 1024 atoms of magnesium (Mg)?
monitta
N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹

n(Mg)=N/Nₐ

m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ

m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
5 0
3 years ago
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