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d1i1m1o1n [39]
3 years ago
11

HELP!!!!!!!!!!! *What are some of the safety concerns for Barium sulfate & Barium Sulfide? Explain

Chemistry
1 answer:
Alex3 years ago
3 0

BaSO₄ is relatively harmless, but BaS is highly toxic.

BaSO₄ is quite insoluble (240 µg/100 mL). It is a <em>mild irritant</em> in cases of skin contact and inhalation. However, it is <em>safe enough</em> that health professionals ask patients to drink a suspension of BaSO₄. The Ba is opaque to X-rays, so it makes the stomach and intestines more visible to radiographers.

BaS is soluble (7.7 g/100 mL). It reacts slowly with water and more rapidly in the acid conditions of the stomach to <em>release H₂S</em>.

BaS + 2HCl ⟶ BaCl₂ + H₂S

An H₂S concentration of 60 mg/100 mL can be <em>fatal within 30 min</em>.

<em>Don’t eat barium sulfide!</em>

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const2013 [10]
I think it is B

Explained answer:
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7 0
3 years ago
Does anyone know the answer to this chemistry question?!? How much HCl is made when 2g of hydrogen reacts? H2+Cl2-&gt;2HCl
jonny [76]
<h3>Answer:</h3>

72.19 g HCl

<h3>Explanation:</h3>

We are given the reaction between hydrogen gas and chlorine gas as;

H₂(g) + Cl₂(g) → 2HCl(g)

Mass of hydrogen gas that reacts is 2 g

We are required to determine the amount of HCl produced

<h3>Step 1: Determine the number of moles of H₂ used </h3>

Moles = Mass ÷ molar mass

Molar mass of Hydrogen gas = 2.02 g/mol

Thus;

Moles = 2 g ÷ 2.02 g/mol

          = 0.99 moles

<h3>Step 2: Moles of HCl produced </h3>

From the reaction equation, 1 mole of hydrogen gas reacts to produce 2 moles of HCl

Therefore, the mole ratio of H₂ : HCl = 1 : 2

Thus, moles of HCl = 0.99 moles × 2

                                = 1.98 moles

<h3>Step 3: Mass of HCl produced </h3>

To calculate mass we multiply the number of moles by molar mass

Molar mass of HCl =36.46 g/mol

Therefore;

Mass of HCl = 36.46 g/mol × 1.98 moles

                    = 72.19 g

Thus, the amount of HCl produced during the reaction is 72.19 g

3 0
2 years ago
If hydrochloric acid is obtained commercially at a concentration of 12.1M, how many milliliters of 12.1M HCl(aq) must be used to
rewona [7]

Answer:

V_1=82.6mL

Explanation:

Hello there!

In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

M_2V_2=M_1V_1

Thus, we solve for the initial volume, V1, as shown below:

V_1=\frac{M_2V_2}{M_1}

And plug in the initial concentration and final concentration and volume to obtain:

V_1=\frac{2000mL*0.500M}{12.1M}\\\\V_1=82.6mL

Regards!

7 0
3 years ago
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givi [52]

Explanation:

In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.86 grams. Determine ...

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