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blagie [28]
2 years ago
6

Which atom has the smallest radius? Sr or I

Chemistry
2 answers:
amid [387]2 years ago
7 0
I think it is hydrogen
sweet [91]2 years ago
5 0

Answer: Strontium has 219 pm

Iodine has 115 pm

Explanation: So the answer is Sr-Strontium

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Josie created a Venn Diagram to determine the differences and similarities between weather and climate. PLSSS HELP I HAVE 15 MIN
DENIUS [597]

Answer:

Climate is long time intervals and weather is short time intervals

Explanation:

4 0
3 years ago
How would you balance this equation<br><br> HCL+ Ba &gt; BaCL2 + H2
vladimir2022 [97]
<span>2 HCl + Ba(OH)2 = BaCl2 + 2 H2O is the answer.</span>
5 0
3 years ago
Read 2 more answers
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo
frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J

Stape 2: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J

The total work is the sum of the two steps

W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J

4 0
3 years ago
How many moles of CO2 are priloduced when 3.5 moles of C2H2 reacts?​
qwelly [4]

Answer:

7mol of CO2

Explanation:

The balanced equation for the combustion of acetylene is

2C_{2} H_{2}+5O_{2}=4CO_{2}+2H_{2}O

Using the equation coefficients, we can find out the number of moles of CO2 produced as follows.

\frac{3.5mol C_{2}H_{2}}{1} *\frac{4CO_2}{2 C_{2}H_{2}} =7molCO_{2}

5 0
3 years ago
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