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Airida [17]
3 years ago
6

How many significant figures in 5241.034?

Chemistry
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

101.44

Explanation:

idk

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I need an answer now please asap and you will me marked brainiest please it's missing I need it now.
Phoenix [80]

Answer:

Just put 1 in the boxes

Explanation:

3 0
2 years ago
Read 2 more answers
A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th
Kisachek [45]

Answer:

pH=4.1

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1

Regards.

3 0
4 years ago
Describe, in terms of electrons, the difference between a calcium atom and a calcium ion
DanielleElmas [232]

The Calcium atom has 2 valence electrons but a Calcium ion will have no electrons because it is a cation ion. Meaning that there are more protons than electrons. Also, the normal calcium symbol is Ca but the calcium ion is Ca²⁺

I hope this wasn't too late

3 0
3 years ago
If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of ai
Gnom [1K]

In accordance with Dalton's Law of multiple proportions

<h3>Further explanation</h3>

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

  • air(O₂) as a limiting reactant(product=14 g)

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

  • air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions

4 0
3 years ago
What is a Fahrenheit??
Scrat [10]

Fahrenheit is a temperature scale

8 0
3 years ago
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