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3 years ago

In parallel circuits different _____________ different branches of the wire?

Chemistry

1 answer:

Helga [31]3 years ago

3 0

I believe that this is the answer :

in parallel circuits, different loads has different branches of wire

good luck

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Determine the percent composition of potassium dichromate, K2Cr2O7.

bulgar [2K]

Answer:

Option E is the correct one

Explanation:

K₂Cr₂O₇ → Potassium dichromate

A ionic salt which its molar mass is 294.18 g/m

It is composed by 2 mol of K, 2 mol of Cr and 7 mol of O

2 mol of K = 78.2 g

2 mol of Cr = 104 g

7 mol of O = 112 g

In 294.2 g of compound we have __78.2 g K __ 104 g Cr ___ 112 g O

Let's calculate the percent

(78.2 / 294.2) .100 = 26.59% K

(104 / 294.2) .100 = 35.35% Cr

(112 / 294.2) . 100 = 38.06 % O

4 0

3 years ago

The carbon-carbon (C―C) bond has an average bond energy of 347 kJ/mol, which is the energy required to break one mole of C―C bon

olganol [36]

Answer:

(3) 345 nm

Explanation:

Given:

Average C-C bond energy = 347 kJ/mol

To determine:

Wavelength of photon that can break a C-C bond

Calculation:

The energy (E) of a photon is related to its wavelength (λ) by the Planck's equation:

$E = \frac{hc}{\lambda }$

where h = Planck's constant = 6.626*10⁻³⁴ Js

c = speed of light = 3*10⁸ m/s

$\lambda = \frac{hc}{E}$

$\lambda =\frac{6.626*10^{-34}Js*3*10^{8}ms^{-1}*6.023*10^{23}mol^{-1}}{347,000Jmol^{-1}}$

λ = 3.45*10⁻⁷ m

Since 1 nanometer (nm) = 10⁻⁹ m

The calculated wavelength corresponds to 345 nm

6 0

4 years ago

Given 2.91 moles of a gas in a 500 milliliter-container, if the temperature is found to be 31 degrees Celsius, what is the press

kari74 [83]

Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K

P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm

3 0

3 years ago

Read 2 more answers

When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is t

taurus [48]

Answer:

The standard enthalpy of formation of HgO is -90.7 kJ/mol.

Explanation:

The reaction between Hg and oxygen is as follows.

$\text{Hg(l)}+\frac{1}{2}{O_{2}\rightarrow \text{HgO(s)}$

From the given,

Molar mass of HgO = 216.59 g/mol

Mass of HgO decomposed = 18.5 g

Amount of heat absorbed = 7.75 kJ

From the reaction,

The standard enthalpy of formation = $+7.75\times\frac{kJ}{18.5 g}\frac{216.59}{1mol} \,\,= +90.7 kJ/mol$

During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.

For the formation of 1 mol of HgO , 90.7 kJ of energy is release

Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol

5 0

3 years ago

Identify the acid and conjugate base pair in the following equation:

Serjik [45]

Answer:

hola no ablo inalgun moderador puede acabar a este otro este me borró respuesta que no debió borrar

7 0

3 years ago