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Fudgin [204]
3 years ago
5

What did you observe in the reaction of phosphoric acid with the freshly cleaned zinc when compared with a piece just taken out

of the bottle
Chemistry
1 answer:
mash [69]3 years ago
6 0

Answer:

The classification is mentioned below for the particular topic.

Explanation:

  • Whether we position 2 different beakers in such a single beaker through one clean edge of zinc-containing H₃Po₄ and another one with unflushed zinc.  
  • The zinc that was washed set to release hydrogen gas way quicker, unlike unventilated zinc.  

⇒      3Zn+2H_{3}Po_{4}\rightarrow Zn_{3}(Po_{4})_{2}+3H_{2}

  • Since fresh zinc complicates the cycle since, as a comparison to polluted zinc, there was little contact with either the reaction.
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Calculate the ratio of effusion rates of cl2 to f2 .
WINSTONCH [101]
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
6 0
3 years ago
How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?
zloy xaker [14]

Answer:

m_{HCl}=36.1gHCl

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl

Regards!

4 0
2 years ago
Part A:
uranmaximum [27]

Answer:

Part A = The mass of sulfur is 6.228 grams

Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams

Explanation:

Part A

Step 1: Data given

A mixture of carbon and sulfur has a mass of 9.0 g

Mass of the product = 27.1 grams

X = mass carbon

Y = mass sulfur

x + y = 9.0  grams

x = 9.0 - y

x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6

(9 - y)*(44.01/12.01) + y(64.07/32.07)

(9-y)(3.664) + y(1.998)

32.976 - 3.664y + 1.998y = 22.6

-1.666y = -10.376

y = 6.228 = mass sulfur

x = 9.0 - 6.228 = 2.772 grams = mass C

The mass of sulfur is 6.228 grams

Part B

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Calculate moles of 1 silver atom

Moles = 1/ 6.022*10^23

Moles = 1.66*10^-24 moles

Mass = moles * molar mass

Mass = 1.66*10 ^-24 moles *107.87

Mass = 1.79 * 10^-22 grams

The mass of 1 silver atom is 1.79 * 10^-22 grams

5 0
2 years ago
Read 2 more answers
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

5 0
2 years ago
Which of the following acts as a catalyst in catalytic converters?
Serhud [2]

Answer:

B. Metal

Explanation:

The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium. Platinum is used as a reduction catalyst and as an oxidation catalyst. Although platinum is a very active catalyst and widely used, it is very expensive and not suitable for all applications.

Hope it helps plz mark brainlist :)

5 0
3 years ago
Read 2 more answers
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