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Dmitry [639]
2 years ago
9

How many protons are in the nucleus of an atom with an atomic number of 15

Chemistry
1 answer:
kykrilka [37]2 years ago
4 0

Answer:

15 protons

Explanation:

The atomic number always equals the number of protons inside the nucleus, so if the atomic number is 15, that means 15 protons are present.

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Lina has worked hard to gather data for her experiments. With her last set of data she received, she is ready to publish her pap
nika2105 [10]

43.8 has 3 significant figures and 1 decimal.

<h3 /><h3>What are significant figures?</h3>

The term significant figures refer to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation.

All zeros that occur between any two non-zero digits are significant. For example, 108.0097 contains seven significant digits. All zeros that are on the right of a decimal point and also to the left of a non-zero digit are never significant. For example, 0.00798 contained three significant digits.

Hence, 43.8 has 3 significant figures and 1 decimal.

Learn more about significant figures here:

brainly.com/question/14359464

#SPJ1

8 0
1 year ago
Why do atoms of nonmetals form anions? Support your answer with an example.
Papessa [141]
Nonmetals form negatively charged ions, or anions. They do this because they need to gain one to three electrons in order to achieve an octet of valence electrons, making them isoelectronic with the noble gas at the end of the period to which they belong.
5 0
3 years ago
How many moles of Carbon are in 3.06 g of Carbon
natta225 [31]

Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

  • Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Since we are converting 3.06 grams to moles, we multiply by that value.

3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

3.06 * \frac{1 \ mol \ C}{12.011 }    

\frac {3.06}{12.011 } \ mol \ C                                

0.25476646 \ mol \ C

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

  • 0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

0.255 \ mol \ C

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0
2 years ago
What is the best explanation of why the iron found in the inner core of the Earth remains in the solid state even though the tem
Musya8 [376]

Answer: b

Explanation: the pressure from gravity in the inner core prevents the iron from melting to a liquid

8 0
3 years ago
An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a
nekit [7.7K]

Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ=  P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂=  P₁ +ρ*v₁²/2 - ρ*v₂²/2  = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂  = P₁ *ρ₁  + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2  = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁  + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) =  (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂=  1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

8 0
3 years ago
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