All categories

3 years ago

How many grams of NH3 are needed to provide the same number of molecules as in 0.55 grams of SF6?

Chemistry

1 answer:

xenn [34]3 years ago

4 0

Above it says the molecular weights are

NH3- 17g/mol and SF6-146 g/mol

Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.

so we would need 0.00582000438 moles of NH3 to have the same number of molecules.

One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:

<span><span>= 17.030519999989988 g / mole * </span>0.00582000438moles</span>

that equals 0.09911770099 grams.

so 0.09911770099 grams is the answer if you rounded, you get about 0.1 grams

You might be interested in

Why does weathering cause erosion

Naily [24]

Answer:Why does weathering cause erosion

Explanation:Organic weathering happens when plants break up rocks with their growing roots or plant acids help dissolve rock. Once the rock has been weakened and broken up by weathering it is ready for erosion. Erosion happens when rocks and sediments are picked up and moved to another place by ice, water, wind or gravity.

7 0

2 years ago

Read 2 more answers

the atom was thought to be indivisible. Explain the significance of Thomson's concept of the electron.

Ludmilka [50]

Answer:

Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup.

4 0

3 years ago

If an individual is eating a diet in which 25% of their ATP production is a result of utilizing carbohydrates, 50% of their ATP

NISA [10]

Answer:

Oxidative phosphorylation proceeds with the formation of energy laden molecules i.e; carbondioxide and water.

Therefore, Total CO₂ production is directly related to VCO₂ = R x VO₂

where, R is the respiratory quotient varing among 0.7 to 1.0 according to the energy intake (ATP) ie 0.25 of the total diet consumed .

VO₂ is, as mentioned above arterial venous oxygen difference = 6.2ml/dl

therefore, VCO₂ = 0.25 x 6.2

= 1.55 ml/dl

ie; VO₂ : VCO₂ = 6.2 : 1.55.

Explanation:

5 0

2 years ago

It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?

Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present = 1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9 = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

#SPJ1

4 0

1 year ago

The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was

Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

$[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}$

$[Ca^{2+}]=0.007118 \ M$

$[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}$

$=0.001994 \ M$

Mass of Ca²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}$

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}$

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0

2 years ago