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IRISSAK [1]
1 year ago
14

zn (s) 2 hcl (aq) -----> zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0

atm, how many g of zn was used at the start of the reaction? (vapor pressure of water at 28oc is 28.3 mmhg).
Chemistry
1 answer:
nadya68 [22]1 year ago
3 0

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn

Read more about zinc;

brainly.com/question/28880469

#SPJ4

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NARA [144]

Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

8 0
3 years ago
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In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is
irga5000 [103]
Answer:
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Explanation:
                   In this reaction first of all HBr approaches the triple bond. A Pi Complex (weak inter-molecular interactions) is formed between the two molecules. And the triple bond attacks the partial positive hydrogen atom creating a negative charge on Bromine along with positive charge on itself (Sigma Complex). In second step the negative Bromide attacks the positive carbon of Butyne.</span>
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3 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

3 0
3 years ago
Read 2 more answers
How many liters of water vapor are in 36.21 g?
Elden [556K]

Answer:

45.02 L.

Explanation:  

  • Firstly, we need to calculate the no. of moles of water vapor.
  • n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
  • We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

<em><u>Using cross multiplication:</u></em>

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0
3 years ago
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