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pychu [463]
3 years ago
15

Predict the size of the astatine atom compared to that of tellurium

Chemistry
1 answer:
PIT_PIT [208]3 years ago
6 0
 <span>Since Astatine is in period 7 and Te is in period 6, I'm going with At because it has 7 energy levels.</span>
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A sample of sulfur hexafluoride gas occupies a volume of 9.10 L at 198ÁC. Assuming that the pressure remains constant, what temp
alexandr1967 [171]
  As we know that
<span>V1/T1 = V2/T2 
V1 = 9.10 L 
T1 = 471 K 
V2 = 2.50 L 
T2 = 2.5 x 471 / 9.10 = 129.3 K 
T2 = 129.3 - 273 =
 -143.6 deg Celsiu
hope it helps</span>
5 0
4 years ago
Read 2 more answers
A 0.500-g sample of KCl is added to 50.0g of water in a calprimeter (Figure 5.12) If the temperature decreases by 1.05C. what is
Sauron [17]

Answer : The reaction is endothermic.

Explanation :

Formula used :

Q=m\times c\times \Delta T

where,

\Delta T = change in temperature = 1.05^oC

Q = heat involved in the dissolution of KCl = ?

m = mass = 0.500 + 50.0 = 50.5 g

c = specific heat of resulting solution = 4.18J/g^oC

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

Q=50.5g\times 4.18J/g^oC\times 1.05^oC

Q=+221.64J

The heat involved in the dissolution of KCl is positive that means as the change in temperature decreases then the reaction is endothermic and as the change in temperature increases then the reaction is exothermic.

Hence, the reaction is endothermic.

8 0
3 years ago
2Al + 6HCl --&gt; 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
9. For each reaction listed, identify the proton donor (or acid) and the proton acceptor(or base).
Talja [164]

Answer:

Explanation:

conjugate acid, based on Brønsted–Lowry acid–base theory, is a chemical compound that is formed by the reception of a proton by a base

a. CH₃COOH + H₂O ⇌ H₃0⁺ + CH₃C00-

Acid <> CH₃COOH

Base <> H₂O

Conjugate acid <> H₃0 +

Conjugate base <>CH₃C00-

b. HCO₃ + H₂O ⇌ H₂CO₃⁻ + OH⁻

Acid <> H₂O

Base <> HCO₃

Conjugate acid <> H₂CO₃⁻

Conjugate base <>OH⁻

C. HNO₃ + SO₄²⁻ ⇌ HSO₄⁻ + NO₃⁻

Acid <>HNO₃

Base <>SO₄²⁻

Conjugate acid <>HSO₄⁻

Conjugate base <>NO₃⁻

A Bronsted acid is reffered to as a proton donor while a Bronsted base is a proton acceptor

3 0
4 years ago
Which of the following represents the least number of molecules?
lozanna [386]

Answer:

A ; 20g of water has the least number of molecules

Explanation:

Here, we want to know which of the options represent the least number of molecules;

To calculate this, we are going to first calculate the number of moles using the formula below;

Number of moles = mass/molar mass

Then from the number of moles, we can calculate the number of molecules by multiplying the number of moles by 6.02 * 10^23 molecules

Let’s proceed;

a. That will be 20/18.02 = 1.1099 moles

1.1099 * 6.02 * 10^23 = 6.68 * 10^23 molecules

b. That will be 77/16.06 = 4.8 moles

4.8 * 6.02 * 10^23 = 2.89 * 10*24 molecules

c. That will be 68/42.09 = 1.62 moles

1.62 * 6.02 * 10^23 = 9.73 * 10^23 molecules

d. That will be 100/44.02 = 2.27 moles

2.27 * 6.02 * 10^23 = 1.37 * 10^24 molecules

e. That will be = 84/20.01 = 4.2 moles

4.2 * 6.02 * 10^23 = 2.53 * 10^24 molecules

From all the values obtained, the lowest is 20g of water

6 0
4 years ago
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