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2 years ago

How many moles FeBr3 are required to generate 275 g NaBr? 2FeBr3 + 3Na₂S → Fe₂S3 +6NaBr

Chemistry

1 answer:

slega [8]2 years ago

8 0

Answer:

0.893mol

Explanation:

n = m ÷ M

= 275 ÷ (23 + 80)

= 2.67 mol

* now use the Mol ratio *

NaBr : FeBr3

6 : 2

2.67 : x

5.67 = 6x

n( FeBr3 ) = 0.893 mol

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Answer:

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You test the pH of the kiddie pools water. You discover thats its pH is 6.3. Is the water too acidic or too basic? how could thi

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3 years ago

Read 2 more answers

What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu?

Xelga [282]

The composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.

<h3>How to convert weight % to atom %?</h3>

The weight percentage of an element can be converted to atom percentage as follows:

C(Zn) = C1A2 / (C2A1) + (C1A2) × 100
C(Cu) = C2A1 / (C1A2) + (C2A1) × 100

Where;

C1 = 30
C2 = 70
A1 = 65.4g/mol
A2 = 63.55g/mol

C(Zn) = {30*63.55} / (70*65.4) + (30*63.55) × 100
C(Cu) = {70*65.4} / (30*63.55) + (70*65.4) × 100

C(Zn) = 29.4 at %
C(Cu) = 70.6 at %

Therefore, the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.

Learn more about atomic percent at: brainly.com/question/10018123

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2 years ago

D: ____________________________________________

Vikki [24]

Answer:

I think the awnser would be c urf mom

Explanation:

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3 years ago

A sample of sulfur dioxide (SO2) is initially at a temperature of 105C, a volume of 15 L, and a pressure of 1.5 atm. If the volu

Agata [3.3K]

The new pressure is 1.08 atm

From the question, we are to determine the new pressure

Using the General gas equation,

$\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

Where P₁ is the initial pressure

V₁ is the initial volume

T₁ is the initial temperature

P₂ is the final pressure

V₂ is the final volume

and T₂ is the final temperature

From the given information,

P₁ = 1.5 atm

V₁ = 15 L

T₁ = 105 C = 105 + 273.15 = 378.15 K

V₂ = 25 L

T₂ = 181 C = 181 + 273.15 = 454.15 K

P₂ = ?

Putting the parameters into the equation, we get

$\frac{1.5 \times 15}{378.15} = \frac{P_{2} \times 25 }{454.15}$

$P_{2}= \frac{1.5 \times 15 \times 454.15}{25 \times 378.15}$

P₂ = 1.08 atm

Hence, the new pressure is 1.08 atm

Learn more on Gas laws here: brainly.com/question/25736513

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2 years ago