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pychu [463]
3 years ago
9

The state of California experiences periodic occurrences of severe drought conditions. Which weather phenomenon most likely caus

es this
Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0
I would say that La Nina is responsible for dry conditions in California and typically it is the year after La Nina that the dryness turns often to severe drought. Drought can be bad for agriculture requiring a lot more irrigation and posing a drain on water resources which decrease also during a drought,
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3 years ago
1. Aluminium reacts with chlorine to produce aluminium chloride. The balanced reaction is:
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Answer:

Explanation:

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3 years ago
In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}

n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{Fe} = 0.5\:mol}

* to n(S)

n_{S} = \dfrac{m_{S}}{MM_{S}}

n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{S} = 0.5\:mol}

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

Fe + S \Longrightarrow FeS

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}

m_{FeS} = n_{FeS}*MM_{FeS}

m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

44 grams of iron sulfide

___________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

4 0
3 years ago
A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

4 0
3 years ago
An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two w
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Answer:

Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.  

Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.  

3 0
3 years ago
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