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Anika [276]
3 years ago
5

Something made of all one material

Chemistry
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

element

Explanation:

an element is something made up of only one type of atom

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The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH
kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:

0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

3 0
2 years ago
Why do you think that the electrons are not shown?
snow_lady [41]
Because there to tiny to see
7 0
2 years ago
(台)what is the answer ​
alukav5142 [94]

Answer:

google chrome

Explanation:

it is the home button on the top left corner

3 0
2 years ago
Scientists need special tools to work in the laboratory. These tools are known as___________
inessss [21]
Instruments

The specific type of instruments depends on the type of laboratory that you're working in: some labs for example use electron microscopes, others use mass spectrophotometers, others use multiplex biochemical analyzers, etc. But very broadly, the specialized tools we use in the laboratory are usually referred to as "instruments"
3 0
3 years ago
Select the correct answer. Given: 2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams
ivann1987 [24]

The reaction produces 2.93 g H₂.

M_r:                        133.34  2.016

        2Al + 6HCl → 2AlCl₃ + 3H₂

<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

7 0
3 years ago
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