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3 years ago

R4te me for 10 points and be honest with me PLEASE

Chemistry

2 answers:

Hoochie [10]2 years ago

8 0

Answer:

id give you and 8 or 9 but you are really pretty tho

Explanation:

dezoksy [38]2 years ago

7 0

i give you an 7 i am being honest i was gonna give you a 6 but then your hair is pretty

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How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?

Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

S1=50%

V2=250ml

S2= 20%

We all know that

V1S1=V2S2

∴V1= V2×S2÷S1

∴V1= V2S2×1/S1

∴V1= 250×20÷50

∴V1= 100ml

6 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

A person's heartbeat is 73 beats per minute. If his heart beats 3.1e9 times in a lifetime, how long does the person live?

tiny-mole [99]

Answer: The person lived for 80.66 years.

Explanation:

Heart beat of the person = 73 beat /min

Number of total heart beats = $3.1\times 10^{9}$

Life span of the person : $\frac{\text{Total heart beats of the person}}{\text{Heart beat per minute}}=\frac{3.1\times 10^{9} beats}{73 beat/min}$

1 year =525600 mins

Life span = $0.0424\times 10^{9} min=80.66 years$

The person lived for 80.66 years.

7 0

2 years ago

What did Thomas Young, Max planck and Albert Einstein contribute towards the nature of light?

Lera25 [3.4K]

The contribution to the nature of light-

Thomas Young - wave nature of light (double-slit experiment)

Max Planck - E = hv

Albert Einstein - a quantum theory of light

Thomas Young proposed the most important double-slit experiment which shows that light acts like a wave and shows the pattern of interferences.

Max Planck proposed that light is proportional to frequency. He gave the equation, E = hv, where E is the energy of light, h is Planck's constant, and v is the frequency.

Albert Einstein proposed the quantum theory of light. He determined that light exists in discrete quanta of energy called photons.

To learn more about the nature of light, visit: brainly.com/question/4423091

#SPJ9

5 0

2 years ago

Help in chem please!!!!!!

IgorLugansk [536]

This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

$1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}$

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0

3 years ago