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3 years ago

What is the standard value for T in the common form of the Nernst equation?

Chemistry

2 answers:

Sveta_85 [38]3 years ago

8 0

The answer to your problem would b b

xenn [34]3 years ago

8 0

The nearest would be B

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How many grams of carbonic acid were produced by the 3.00 g sample of NaHCO

stellarik [79]

I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3

mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3

7 0

3 years ago

Balance the following equation. Then determine the ratio for the products KCl and O2 generated during the decomposition of potas

drek231 [11]

The answer is 2:3 for plato users

6 0

2 years ago

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What is the total number of liters of NH3 formed when 20 liters of N2 reacts completely

goblinko [34]

Answer:

N2 + H2 ----------》NH3

On balancing it

N2. + 3.H2------->2.NH3

( 1 mol) (3 mol) (2 mol)

1 mol of nitrogen reacts with 3 mol of hydrogen to give 2 mol of ammonia.

Likewise,

20 litres of nitrogen reacts with 60 litres of hydrogen to give 40 litres of Ammonia.

Hence, the answer is 40 Litres.

5 0

3 years ago

ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y

aleksandr82 [10.1K]

Answer:

The expression to calculate the mass of the reactant is $m = \frac{1.080kJ}{31.2kJ/g}$

Explanation:

The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant. To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

$1,080J.\frac{1kJ}{10^{3}J } =1.080kJ$

Then,

$1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}$

5 0

3 years ago

Read 2 more answers

How many moles do you have in 37.3 g of Co(CrO4)3

shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

37.3 g Co(CrO₄)₃

Step 2: Identify Conversions

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

Step 3: Convert

$37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} )$ = 0.091662 mol Co(CrO₄)₃

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0

2 years ago