Which of the following sequences of elements would all have four valence electrons?

In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area

storchak [24]

First we must write a balanced chemical equation for this reaction

$Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)$

The mole ratio for the reaction between $HNO_3$ and $Na_2CO_3$ is 1:2. This means 1 moles of $Na_2CO_3$ will neutralize 2 moles $HNO_3$ . Now we find the moles of each reactant based on the mass and molar mass.

$2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3$

$2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3$

$\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}$

The $Na_2CO_3$ was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.

4 0

3 years ago

How many moles are equal to 89.23g of calcium oxide, CaO?

Katen [24]

1.59moles

Explanation:

Mass of CaO = 89.23g

Unknown

Number of moles = ?

Solution:

The mole is a unit of measurement in chemistry used to delineate the number of particles an atom contains.

A mole of a substance contains the avogadro's number of particles.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CaO = 40 + 16 = 56g/mol

Number of moles = $\frac{89.23}{56}$ = 1.59moles

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

8 0

3 years ago

How many moles of glucose (C6H12O6) are in 5.0 liters of a 2.5 M C6H12O6 solution?

Vaselesa [24]

5*2.5 = 12.5 ≈ 13 moles

3 0

3 years ago

Read 2 more answers

Which of the following statements best describes the relationships between carrying captivity in population size

Ierofanga [76]

The correct answer is letter a, whereas, carrying capacity in captivity increases population size. It is because the carrying captivity is the one responsible for having a maximum population size in which will help in sustaining the necessities that the species need in the environment in which makes it responsible for the population size to increase depending on its capacity. The correct answer is letter a.

4 0

3 years ago

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers