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3 years ago

How do you balance the equation in number one?

Chemistry

1 answer:

AleksAgata [21]3 years ago

3 0

____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O

To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.

To start off, you wanna know the number of atoms in each element on both sides, so take it apart:

[reactants] [product]
Na- 1 Na- 2

N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)

O- 4 O- 7 (same reason as above)

Pb- 1 Pb- 1

Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:

2NaNO3

Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:

Na- 2 Na- 2

N- 2 N- 2 (it balanced out)

O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)

Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).

Now to put it back together, it will look like this:

2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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3 years ago

Mannoheptulose is a sugar found in avocados. If each C-C bond contains 76 kcal of energy and each C-H bond contains 91 kcal, how

nadya68 [22]

Answer:

Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal

Explanation:

The molecular formula of mannoheptulose is C₇H₁₄O₇.

The structure is as shown in the attachment below.

Number of C-C bonds present in mannoheptulose = 6

Number of C-H bonds present in mannoheptulose = 8

Since the each C-C bond contains 76 Kcal of energy,

Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal

Also, since each C-H bond contains 91 Kcal of energy;

amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal

Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal

6 0

3 years ago

From the lists of available reagents select the one(s) you would use to in a preparation of acetophenone (phenyl methyl ketone)

VARVARA [1.3K]

Answer:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

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3 years ago

Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at

WITCHER [35]

Answer: The volume of given amount of ethanol at this temperature is 159.44 mL

Explanation:

Specific gravity is given by the formula:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

$0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL$

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

$\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL$

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0

3 years ago

How does this process appear in the atmosphere?

Mariulka [41]

D.The transferring of thermal energy from the sun to the earth!

8 0

3 years ago

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