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3 years ago

What is the longitude of the international date line

Chemistry

1 answer:

sveta [45]3 years ago

8 0

Answer:

The International Date Line passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south

line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich

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Please help, answer questions 3-7

tatyana61 [14]

I don’t know about 3 but I know that 4 is solid, 5 is you’ll feel the vibration first, 6 is you’ll see the lightning first because the speed of light is faster than the speed of sound and 7 is sound waves need a medium to travel and light waves don’t, sound waves are longitudinal and light waves are transverse, and sound waves are mechanical waves while light waves are electromagnetic waves

3 0

3 years ago

Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?

Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

Which two characteristics describe all animals

Maru [420]

All animals can be dangerous and they would fight for their family. (This might be wrong)

5 0

3 years ago

What is the difference in energy between an electron in 1s and an electron in 4s.

Andre45 [30]

Answer:

Explanation:

An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.

5 0

3 years ago