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2 years ago

How would you know if this combination is likely to be found in dirt? Please explain!

Chemistry

2 answers:

Natasha2012 [34]2 years ago

5 0

No at least I don’t think there would be

Elan Coil [88]2 years ago

4 0

No it is not likely. That is a ratio of 10:4 N^14 and N^15 which doesn’t work. It needs a higher amount

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a laboratory procedure calls for making 510.0 mL of a 1.6 M KNO3 solution. How much KNO3 in grams is needed

barxatty [35]

Answer:

82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.

Explanation:

Since molarity is the number of moles of solute that are dissolved in a given volume, calculated by dividing the moles of solute by the volume of the solution, the following rule of three can be applied: if in 1 L (1,000 mL) of KNO₃ there are 1.6 moles of the compound present, in 510 mL how many moles will there be?

$moles=\frac{510 mL*1.6 moles}{1000 mL}$

moles= 0.816

Being the molar mass of the elements:

K: 39 g/mole
N: 14 g/mole
O: 16 g/mole

So the molar mass of the compound KNO₃ is:

KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

Now I can apply the following rule of three: if in 1 mole of KNO₃ there are 101 g, in 0.816 moles how much mass is there?

$mass=\frac{0.816 moles*101 grams}{1 mole}$

mass= 82.416 grams

82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.

4 0

2 years ago

How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C

yaroslaw [1]

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m=21 g
ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

4 0

3 years ago

The universe contains billions of galaxies, each containing billions of stars. True or false

Tatiana [17]

True but no one know Exactly how many stars there are

8 0

3 years ago

Read 2 more answers

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

PLEASE HELP ASAP!!!!!!

WINSTONCH [101]

The immensely strong heat creates conviction cells to create movement as it follows the currents.

7 0

3 years ago