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3 years ago

8

Hydrogen gas explosive within range of 4%-75% v/v. Assuming that each student in your class produces 6 L of H2 and that the lab

is not ventilated, is danger of an explosion? Show your calculations to back up your answer.

1 answer:

SpyIntel [72]3 years ago

3 0

To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.

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Akosua has observed that anytime she drinks sobolo, her stool appears dark green. On one occasion

Anna007 [38]

Answer:

i. Sobolo is a Ghanian drink that is produced from red hibiscus flower that has an average pH of 6.7

It contains cyanidin and anthocyanins, which is a red pigment that is red in an acidic medium and changes green when introduced in a basic medium that has a high pH

The pH at the rectum of the digestive system = 5 to 8 (Slightly basic)

Therefore, what made the stool of Akosua green is that the sobolo drink changes to green in basic solution

ii. The stool which appeared green because she took sobolo turn into bright red upon mixing with the acidic WC water because of the presence of anthocyanins in sobolo, it turns red in an acidic medium

iii. Sobolo which turns green, or blue in a basic medium and red in an acidic medium can be used as a litmus solution to test the pH of a given substance

Explanation:

Sobolo or soobolo in Ghana is a name for the Hibiscus tea or tisane, which is made from calyces of the hibiscus plant, and has a sour (tangy) taste and appears bright red in color

8 0

3 years ago

Name the following ionic compounds, keeping in mind that a transition metal cation must include its charge. ____________________

Lilit [14]

Answer :

(a) The name for $TiO_2$ is titanium(IV) oxide.

(b) The name for $BaCl_2$ is barium chloride.

(c) The name for $CuCl_3$ is copper(III) chloride.

(d) The name for $KI$ is potassium iodide.

(e) The name for $SrCl_2$ is strontium chloride.

(f) The name for $CuBr_2$ is copper(II) bromide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive ion is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) $TiO_2$

$TiO_2$ is an ionic compound because titanium element is a metal and oxygen element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on tin is (+4) and the the charge on oxygen is (-2). Thus, the name for $TiO_2$ is titanium(IV) oxide.

(b) $BaCl_2$

$BaCl_2$ is an ionic compound because barium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on barium is (+2) and the the charge on chlorine is (-1). Thus, the name for $BaCl_2$ is barium chloride.

(c) $CuCl_3$

$CuCl_3$ is an ionic compound because copper element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+3) and the the charge on chlorine is (-1). Thus, the name for $CuCl_3$ is copper(III) chloride.

(d) $KI$

$KI$ is an ionic compound because potassium element is a metal and iodine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on potassium is (+1) and the the charge on iodine is (-1). Thus, the name for $KI$ is potassium iodide.

(e) $SrCl_2$

$SrCl_2$ is an ionic compound because strontium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on strontium is (+2) and the the charge on chlorine is (-1). Thus, the name for $SrCl_2$ is strontium chloride.

(f) $CuBr_2$

$CuBr_2$ is an ionic compound because copper element is a metal and bromine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+2) and the the charge on bromine is (-1). Thus, the name for $CuBr_2$ is copper(II) bromide.

3 0

3 years ago

2. As NH4OH is added to an HCl solution, the pH of the solution

chubhunter [2.5K]

Answer:

c

Explanation:

Nh4OH + HCL ---> NH4Cl + H3O

so ph decreases as H3O increases

and OH also decreases

5 0

3 years ago

Read 2 more answers

How did Wegener use similar mountain ranges and areas made of certain types of rock found in Africa and South American as eviden

Effectus [21]

Answer:

the Germany scientist Alfred Wagner describes fold mountains that are found in The world are the same also geological similarity that occur between this continent are the same for example the rock that are found in brazil are the same of that of south africa.

4 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago