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noname [10]
3 years ago
10

A cleaning bottle contains 83.1 g of ammonia. How many molecules of ammonia are in the bottle?

Chemistry
1 answer:
Tasya [4]3 years ago
4 0

Answer:

2.94 x 10^2^4

Explanation:

First we need to find out how many moles of ammonia there are, using the formula: Mass = mr x moles.

We know the mass is 83.1g, now we need to find the mR of ammonia - NH3.

N = 14, H = 1, so 14 + (3x1) = an mr of 17.

Moles = mass/ mr = 83.1/17 = 4.8882

Now we can multiply the moles by avogadro's constant to find the number of molecules:

4.8882 x (6.02 x 10^2^3 ) = 2.94 x 10^2^4 molecules of ammonia

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A sailboat is pushed by a steady wind with a force of 4000 N. Its acceleration is 2
AveGali [126]

Answer:

<h2>2000 kg</h2>

Explanation:

The mass of the sailboat can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{4000}{2}  \\

We have the final answer as

<h3>2000 kg</h3>

Hope this helps you

6 0
3 years ago
A gas has a volume of 25.6 L at a temperature of 278 K. What will be the new temperature, in C, if
Inessa05 [86]
<h2>Hello!</h2>

The answer is: -97.37° C

<h2>Why?</h2>

According to the Charles and Gay-Lussac's Law we have that:

\frac{V}{T}=k

Where:

<em>v</em> is the volume of the gas

<em>t</em> is the temperature of the gas

<em>k </em>is the proportionality constant

From the Gay-Lussac's Law we also have the following relation:

\frac{V1}{T1}=\frac{V2}{T2}

Since we need to find the new temperature (T2) we can use the last equation:

\frac{V1}{T1}=\frac{V2}{T2}\\\\T2=\frac{V2*T1}{V1}=\frac{16.2L*278K}{25.6L}=175.92K

We are asked to find the temperature in Celsius degrees, so, we must convert the result (in K) to Celsius degrees:

Celsius=K-273.15=175.92-273.15=-97.37

So, the temperature is -97.37° C

Have a nice day!

7 0
3 years ago
How can you determine the quantities of reactants and products in a chemical reaction
prohojiy [21]

Use the balanced equation.

5 0
3 years ago
Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip
erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0
3 years ago
What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus
Nastasia [14]

Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

Explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

2P+\frac{5}{2}O_2\rightarrow P_2O_5

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

Best regards!

5 0
3 years ago
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