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sdas [7]
4 years ago
14

A lithium atom that has lost an electron comes near a chlorine atom that has gained an electron. What happens?

Chemistry
2 answers:
lina2011 [118]4 years ago
8 0
C. They form an ionic compound LiCl
steposvetlana [31]4 years ago
8 0

They form an ionic compound

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Population from the phylogenetic tree?
bogdanovich [222]
Mark Brainliest please

Answer :

In my opinion , answer is A
4 0
3 years ago
5
Archy [21]

Answer:

<h3>The answer is option C</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of liquid = 15 mL

density = 2.5 g/mL

We have

mass = 15 × 2.5

We have the final answer as

<h3>37.5 g</h3>

Hope this helps you

6 0
3 years ago
When the glucose transport maximum is reached, _______.?
n200080 [17]
Glucose is extracted in the urine and all of the glucose is reabsorbed. 
8 0
3 years ago
A researcher dilutes 30.0 ml of ethanol to 300.0 ml with distilled water. what is the percentage concentration by volume of the
Len [333]

Answer: 9.09 %

Explanation:

To calculate the  percentage concentration by volume, we use the formula:

\text{Volume percent of solution}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100

Volume of ethanol (solute) = 30 ml

Volume of water (solvent) = 300 ml

Volume of solution= volume of solute + volume of solution = 30+ 300 = 330 ml

Putting values in above equation, we get:

\text{Volume percent of solution}=\frac{30}{30+300}\times 100=9.09\%

Hence, the volume percent of solution will be 9.09 %.

3 0
3 years ago
Read 2 more answers
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
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