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3 years ago

HELP WITH THIS ASAP PLEASE!!! What is the mass in grams of 8.42 x 10^22 atoms of sulfur?

Chemistry

1 answer:

V125BC [204]3 years ago

5 0

Answer:

Answer:

see explanation and punch in the numbers yourself ( will be better for your test)

Explanation:

If you are given atoms you need to divide by Avogadro's number 6.022x10^23

then you will have moles of sulfur-- once you have moles multiply by the molar mass of sulfur to go from moles to grams

mm of sulfur is 32 g/mol

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Which type of electromagnetic wave has the greatest frequency?(1 point)

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c. x-rays

My answer is that x-rays or gamma rays have the greatest (or highest) frequency waves.

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3 years ago

In a certain variety of plants, red flowers appear in majority of plants and blue flowers appear only in a few plants. Which sta

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3 years ago

Acetylene, c2h2, has a standard enthalpy of formation, δh° = 226.7 kj/mol, and a standard entropy change for its formation from

VLD [36.1K]

ΔG⁰ = ΔH⁰ - T ΔS⁰

ΔG⁰ : Standard free energy of formation of acetylene

ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)

ΔS⁰ : Standard entropy change (58.8 J / K. mol)

T : Temperature 25°C = 298 K (room temperature)

ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol

5 0

3 years ago

Read 2 more answers

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .

Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:

$m_{NaOH} = 10^{12.00 - 14.00}\cdot 39.997 g/mol\cdot 0.100 L = 0.0400 g$

7 0

3 years ago