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Amiraneli [1.4K]
3 years ago
9

How many atoms of potassium, K, are in K3PO4

Chemistry
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

1

Explanation:

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Octopus and squids breathe through
svet-max [94.6K]

<em>Octopus and squids breathe</em> <em>like </em><em>fishes </em><em>they </em><em>breathe </em><em>from </em><em>gills </em>

<em>so </em><em>even </em><em>octopus</em><em> and</em><em> squids</em><em> </em><em>breathe </em><em>through </em><em>gills </em><em>too.</em>

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4 0
3 years ago
During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.
Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL

Putting values in above equation, we get:

1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

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

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0
3 years ago
Help name this, it’s chemistry
vova2212 [387]

Answer:

I believe it's

4-floro-5,6,8-Tribromo-3,3,7-trimethylnonane

5 0
3 years ago
Which experimental setup would let a student investigate the connection between kinetic energy and temperature? (1 point) placin
Maslowich

Answer:

The correct option is;

Placing one drop of food coloring in a cup with 60 ml of water at 10°, placing one drop of food coloring in a second cup with 60 ml of water at 40°C

Explanation:

The experimental setup that would allow the student investigate the connection between kinetic energy and temperature should be made up of the following characteristics

1) The constant terms for the experiment should be defined, which in this case are

a) The volume of the water which is 60 ml in both subjects of the experiment

2) The definition of the variable that produces the effect that is being monitored, which is the use of the different temperatures in the two experimental subjects

3)The environmental limits of the experiment, which is the water and the food coloring used

5 0
3 years ago
A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is
Vlada [557]

Answer:

c=0.127\ J/g^{\circ} C

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

Q=mc\Delta T

c is specific heat of the metal

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C

So, the specific heat of metal is 0.127\ J/g^{\circ} C.

4 0
3 years ago
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