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3 years ago

Im gonna ride it do it just how you like it and after that let's do it?

Chemistry

2 answers:

Reika [66]3 years ago

7 0

Answer:

the hell?

Explanation:

Liono4ka [1.6K]3 years ago

4 0

Dababy............................

You might be interested in

A metallurgist has one alloy containing 21% aluminum and another containing 42% aluminum. how many pounds of each alloy must he

Naily [24]

Answer is: 7.8 lb of 21% aluminum and 33.2 ib of 42% aluminum.

ω₁ = 21% ÷ 100% = 0.21.
ω₂ = 42% ÷ 100% = 0.42.
ω₃ = 38% ÷ 100% = 0.38.
m₁ = ?.

m₂ = ?.
m₃ = m₁ + m₂.
m₃ = 41 pounds.

m₁ = 41 lb - m₂.
ω₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.

0.21 · (41 lb - m₂) + 0.42 · m₂ = 0.38 · 41 lb.

8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.

0.21m₂ = 6.97 lb.

m₂ = 6.97 lb ÷ 0.21.

m₂ = 33.2 lb.

m₁ = 41 lb - 33.2 lb.

m₁ = 7.8 lb.

8 0

3 years ago

A book is pushed off the edge of a table. At what point during the fall will the potential and kinetic energy be equal?

vfiekz [6]

Answer:

KE = PE at half the table Height:

Explanation

AT ANY POINT IN THE BOOK'S FALL,

TOTAL E = PE +KE

THE TOTAL E IS CONSTANT

Before the book is pushed off, the total energy is potential

TOT E=PE =MGH

BEFORE THE BOOK HITS THE GROUND, THE TOTAL E IS KINETIC

TOT=KE = MVXV/2

WHEN KE = PE

KE+PE = MGH (STARTING ENERGY SINCE E IS CONSERVED)

OR PE+ PE = MGH

OR MGH' + MGH' =MGH

OR 2H' =H

H' (NEW HEIGHT) =H/2

5 0

3 years ago

The most active of all the chemical elements is a halogen known as

m_a_m_a [10]

Fluorine! It's highly reactive, since it has 7 electrons and wants an extra electron to fill its valence shell.

4 0

3 years ago

Read 2 more answers

Pls Help! 100 Points!!!!!

Katen [24]

Answer:

The answer is c,F has a smaller radius than F− because F− has an additional energy shell.

Explanation:

if correct mark brainliest

6 0

2 years ago

Read 2 more answers

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

3 years ago