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2 years ago

The correct answer(reported to the proper number of significant figures)to the following is

Chemistry

1 answer:

LenKa [72]2 years ago

8 0

to the proper number of significant figures) to the following? (12.67+19.2)(3.99)/(1.36+ 11.366).

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If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above

jasenka [17]

Answer : The density of an object is, $2g/cm^3$

Solution : Given,

Mass of an object = 60 g

Volume of an object = $30cm^3$

Formula used :

$\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

Now put all the given values in this formula, we get the density of an object.

$\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3$

Therefore, the density of an object is, $2g/cm^3$

6 0

2 years ago

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

Answer: The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

Explanation:

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

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

For calculating the mass of carbon:

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

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

For calculating the mass of sulfur:

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

For calculating the mass of nitrogen:

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

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 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

2 years ago

Iodine has a lower atomic weight than tellurium (126.90 for I, 127.60 for Te) even though it has a higher atomic number (53 for

andrew11 [14]

Answer and Explanation:

Iodine have lower atomic mass than tellurium even though the atomic number of iodine is more than the atomic number of tellurium

This is because the atomic weight of any element is the sum of number of proton and number of neutron, even though the number of proton in iodine is more so but the number of neutron is less as compared to tellurium which makes the tellurium of high atomic mass

6 0

2 years ago

Read 2 more answers

How many ions are there in 0.187 mol of Na+ ions

kondaur [170]

Best Answer
1 mole of a substance contains 6.022x10^23 "units" of that substance.

So 0.187 mol of Na+ is 1.13x10^23 ions (6.022x10^23 x 0.187).

4 0

2 years ago

Read 2 more answers

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

AveGali [126]

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d. withdraws electrons inductively

Explanation:

a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

3 0

2 years ago