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3 years ago

What is mass in science?

Chemistry

2 answers:

Musya8 [376]3 years ago

6 0

Its like weight but not. mass is the amount of matter in a substance.

algol133 years ago

3 0

<span>a coherent, typically large body of matter with no definite shape. may i have brainliest </span>

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What are 3 elements similar to silver

sesenic [268]

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3 0

3 years ago

Read 2 more answers

3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96

arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em /> $\tt W=\dfrac{e.i.t}{96500}$ <em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

$\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g$

8 0

3 years ago

In the world of chemistry, a 'mole' is a number of something... a very large number of something.

kipiarov [429]

Given:

No of atoms present= 8.022 x 10^23 atoms

Now we know that 1 mole= 6.022 x 10^23 atoms

Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.

Number of moles

= 8.022 x 10^23/6.022x 10^23

=1.3 moles.

Hence we have 1.3 moles present.

3 0

3 years ago

a chemist uses hot hydrogen gas to convert chromium (iii) oxide to pure chromium. how many grams of hydrogen are needed to produ

MA_775_DIABLO [31]

Answer:

Explanation:

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6 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

3 years ago