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

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)
Hope this helps :3

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:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of 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: