Which is another word for 10 meters in the metric system?

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

When heated, calcium hydroxide and ammonium chloride react to produce ammonia gas, water vapor, and solid calcium chloride.

NNADVOKAT [17]

Answer:

13.73g

Explanation:

mass of reactants = mass of products.

Mass reactants = 5.00 g + 10.00 g = 15.00 g

Mass products = 1.27g + mass of ammonia and water vapor

Mass of ammonia and water vapor

15.00g – 1.27 g = 13.73 g

3 0

3 years ago

Which of the following is a chemical property of water? It ___________.

lbvjy [14]

The answer is A beautiful

4 0

3 years ago

PLEASE!!! I NEED HELP ASAP!!

Lana71 [14]

W=m₁/m₀=2^(-t/T)

t=4.6·10⁹ years
T=5·10¹⁰ years

w=2^(-4.6·10⁹/5·10¹⁰)

w=0.9382
w=93.82%

7 0

3 years ago

Read 2 more answers

Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio

Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

50.48 % O = 50.48 grams

49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0

3 years ago