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ANTONII [103]
3 years ago
10

How many grams are present in 0.885 moles of manganese?

Chemistry
1 answer:
pshichka [43]3 years ago
4 0
A. 48.6g
PLS MARK BRAINLEIST IF IM RIGHT
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Why is it logical to assume that they hydrogen ion concentration in an aqueous solution of a strong monoproctic acid equals the
natka813 [3]
It is logical to assume that they hydrogen ion concentration in an aqueous solution of a strong monoproctic acid equals the molarity of the acid because ions which are charged,for example ammonium ion (NH4+), which can be derived by the addition of a proton to a molecular base.
3 0
3 years ago
How many moles of Ag will be produced from 16.0 g of Cu, assuming AgNO3 is available in excess
madreJ [45]

Answer:0.504

Explanation:

4 0
3 years ago
The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur
anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × 10^{7}

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × 10^{7}

ΔG° = -2494.2㏑ 1.7 × 10^{7} = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

Learn about formation constant

brainly.com/question/14011682

#SPJ4

8 0
1 year ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?
RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency 1.0 \times {10^{15}}\;{\text{Hz}}  is  \boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

{\mathbf{E=h\nu }}                                   ......(1)

Here,

{\text{E}} is the energy.

h is the Plank’s constant.

\nu is the frequency.

The frequency of UV rays is 1.0 \times {10^{15}}\;{\text{Hz}} or 1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}

The value of Planck’s constant is 6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}} .

Substitute these values in equation (1)

\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is {\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}} .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

7 0
3 years ago
Read 2 more answers
A motorcycle travel at 6.0 m/s. After 3.0 second, the motorcycle travel at 15.0 m/s. Which of the following was the average acce
Lerok [7]

Hello!

\large\boxed{a = 3m/s^{2}}

Use the following equation to solve for the average acceleration of the motorcycle:

a = \frac{v_{f}-v_{i}}{t}

Plug in the given final, initial velocities, and the time:

a = \frac{15-6}{3}\\\\a = \frac{9}{3}\\\\a = 3m/s^{2}

7 0
3 years ago
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