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3 years ago

What happens to the velocity when you apply a constant force to an object? *

Chemistry

1 answer:

Degger [83]3 years ago

8 0

Answer: it is part of newtons second law. so the object to your question would accelerate

Explanation: newtons second law

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Which is a series circuit and a parallel circuit

kirza4 [7]

Explanation:

A is a series circuit

B is parallel circuit

8 0

3 years ago

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Consider the following reaction:

Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

$SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Initial conc. $2.4\times 10^{-2}$ 0 0

At eqm. $(2.4\times 10^{-2}-x)$ x x

As we are given:

Concentration of $Cl_2$ at equilibrium = $1.3\times 10^{-2}M$

That means,

$x=1.3\times 10^{-2}M$

The expression for equilibrium constant is:

$K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$

Now put all the given values in this expression, we get:

$K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}$

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}$

$K_c=0.0154$

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0

3 years ago

HELPPPPP PLEASSEEEEE

timurjin [86]

Answer:

Dalton was first with the modeling
They soon followed so the answer is D: rutherford and bohr modeled their experiments

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3 years ago

What is the electronic configuration of Au3+?

Anettt [7]

Answer:

$AU^{3+} : [Rn] 5f^3$

Explanation:

Writing electronic configuration of any element you should know atomic number of that element ,

and also electrons are filling according to their energy level and first electron is filled in the lower energy orbital

and it follows n+1 rule if n+1 is same for two orbital electron will go first in the lowest value of n.

writing electronic configuration of ion can be done like first for their neutral atom and then add or remove electron it will make things easy because there are also some eception case their you may do wrong.

$AU : [Rn] 5f^3 6d^1 7s^2$

remove three electron from outer most shell of AU

$AU^{3+} : [Rn] 5f^3$

4 0

3 years ago

A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If t

11Alexandr11 [23.1K]

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

8 0

3 years ago

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