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2 years ago

What is the potential difference across a light bulb when there is 32 A of current and 8 Ω of resistance from the filament.

Chemistry

1 answer:

evablogger [386]2 years ago

7 0

Answer:

force ε: the increase in potential energy per unit charge provided by the chemical reactions in the battery. And the voltage difference across the light bulb is: ∆V = I R = (0.06 A)(20 Ω) = 1.2 V ▪If we measure the voltage difference across the battery or the light bulb, we will get 1.2 V.

Explanation:

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A 3.00 g mass of compound x was added to 50.0 g of water and it is found that the freezing point has decreased by 1.25 degrees c

zloy xaker [14]

Answer:

yes

Explanation:

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3 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows:

Nezavi [6.7K]

The theoretical yield of urea : = 227.4 kg

<h3>Further explanation</h3>

Given

Reaction

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

128.9 kg of ammonia

211.4 kg of carbon dioxide

166.3 kg of urea.

Required

The theoretical yield of urea

Solution

mol Ammonia (MW=17 g/mol)

=128.9 : 17

= 7.58 kmol

mol CO₂(MW=44 g/mol) :

= 211.4 : 44

= 4.805 kmol

Mol : coefficient of reactant , NH₃ : CO₂ :

= 7.58/2 : 4.805/1

=3.79 : 4.805

Ammonia as limiting reactant(smaller ratio)

Mol urea based on mol Ammonia :

=1/2 x 7.58

=3.79 kmol

Mass urea :

=3.79 kmol x 60 g/mol

= 227.4 kg

3 0

2 years ago

Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

2 years ago

Which of the following is composed of many cells?

vodomira [7]

Answer:

Bacteria

Explanation:

8 0

3 years ago

Before you touch an electrical switch, plug, or outlet

IceJOKER [234]

Answer is A bc you can get electrocuted

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3 years ago

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