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Semenov [28]
2 years ago
8

What happens to the density of group 1 when you go down the group

Chemistry
1 answer:
OLga [1]2 years ago
4 0

the density increases down the group.

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A 250ml glass of orange juice contains 22 grams of sugar. how much sugar is in a 2 liter (2.500ml) bottle of orange juice?
Maslowich

220 grams of sugar would be in 2 liters of orange juice

5 0
3 years ago
Completely reacting 150.0 g of a substance with oxygen releases 395.1 J of energy. How much energy would be released if 450.0 g
MrMuchimi
To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ

x = 450*395.1/150 = 1185,3J

450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(<span>kiloJoule</span>)
4 0
3 years ago
Read 2 more answers
How do you know anything?
r-ruslan [8.4K]
By studying and learning by using your brain to accomplish new goals. :)
(if that makes sense)
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6 0
3 years ago
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0
3 years ago
A metal M forms the oxide M2O. When 0.890 grams of M reacts with pure oxygen, 0.956 grams of M2O form. Write the balanced equati
Alla [95]

The molar mass of M is  0.225g/mol and the element M is Hydrogen

If a metal M combines with an oxygen element to form the oxide, M_2O then the chemical reaction will be expressed as:

4M + O_2 -> 2M_2O\\

This shows that 4 moles of an unknown element M react with the oxygen element to produce the oxide M_2O

Given the following parameters

Mass of M = 0.890 grams

Mass of M_2O = 0.956 grams

Get the molar mass of M:

Molar mass = Mass/number of moles

Molar mass = 0.890/4

Molar mass = 0.225g/mol

Hence the molar mass of M is  0.225g/mol and the element M is Hydrogen

Learn more here: brainly.com/question/6996520

5 0
2 years ago
Read 2 more answers
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