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3 years ago

Which object converts sunlight into sugars

Chemistry

2 answers:

tatuchka [14]3 years ago

8 0

Plants

do. Good luck with whatever you are doing :)

Burka [1]3 years ago

3 0

Chlorophyll or chloroplast depending on how in depth your teacher wants the answer to be.

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If liquid a is more volatile than liquid b, which is also true? hints if liquid a is more volatile than liquid b, which is also

7nadin3 [17]

liquid A has a lower boiling point than liquid B

7 0

3 years ago

Read 2 more answers

chlorine reacts with fluorine to form flowing monofluoride write an equation for this reaction including the state symbols are t

vovangra [49]

Cl +F = Cl2F

Hope this helped

5 0

2 years ago

How many li are in 2.5 moles of li

NARA [144]

Answer:

15.06 × 10²³ atoms of Li

Explanation:

Given data:

Number of moles of Li = 2.5 mol

Number of toms of Li = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 2.5 mol of Li:

1 mole of lithium = 6.022 × 10²³ atoms of Li

2.5 mol × 6.022 × 10²³ atoms of Li / 1 mol

15.06 × 10²³ atoms of Li

8 0

3 years ago

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th

zaharov [31]

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

E°/V

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

E°/V

2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74

3Fe ⇌ 3Fe²⁺ + 6e-; -0.41

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF}\ln Q$

(a) Data

E° = 0.33 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298 K

z = 6

F = 96 485 C/mol

(b) Calculations:

$Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}$

7 0

3 years ago

A sample of Cd(OH)2 is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Cd +2 = 1.7 x 10 -5M

Archy [21]

Answer:

$Ksp=2.0x10^{-14}$

Explanation:

Hello there!

In this case, given the solubilization of cadmium (II) hydroxide:

$Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)$

The solubility product can be set up as follows:

$Ksp=[Cd^{2+}][OH^-]^2$

Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:

$Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}$

Regards!

3 0

3 years ago