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andrew11 [14]
3 years ago
9

What is the percent cr, by mass, in the steel sample? express your answer numerically as a percentage?

Chemistry
2 answers:
djverab [1.8K]3 years ago
8 0
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−

Assuming all the Cr is contained in the BaCrO4 at the end. 
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
lbvjy [14]3 years ago
3 0

3.40g example of the steel used to deliver 250.0 mL Solution containing Cr2O72−  

Expecting all the Cr is contained in the BaCrO4 toward the end.  

(0.145 g BaCrO4)/(253.3216 g BaCrO4/mol) x (250.0 mL/10.0 mL) x (1 mol Cr/1 mol BaCrO4) x (51.99616 g Cr/mol/(3.40 g)  

= 0.219 = 21.9% Cr  

Further Explanation:  

Chromium (cr):  

Chromium is a compound component with the image Cr and nuclear number 24. It is the primary component in gathering 6. It is a steely-dim, radiant, hard and weak progress metal. Chromium is the principle added substance in tempered steel, to which it includes hostile to destructive properties.  

Characterization of chromium:  

Classification: Chromium is a progress metal  

Color:                                  silver-dark  

Nuclear weight:              51.996  

State:                                        solid  

Softening point:               1907 oC, 2180 K  

synthetic properties of chromium:  

Chromium is a radiant, fragile, hard metal. Its shading is silver-dark and it tends to be profoundly cleaned. It doesn't discolor in air, when warmed it borns and structures the green chromic oxide. Chromium is unsteady in oxygen, it quickly creates a meager oxide layer that is impermeable to oxygen and secures the metal underneath.

Subject: chemistry

Level: High School

Keywords: Chromium (cr), Characterization of chromium, synthetic properties of chromium.

Related links:  

Learn more about evolution on

brainly.com/question/5325748

brainly.com/question/7035326

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strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get:

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

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