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solniwko [45]
3 years ago
6

If the density of a certain spherical atomic nucleus is 1.0 _ 1014 g cm_3 and its mass is 2.0 _ 10_23 g, what is its radius in c

m? I need to understand how to do it not only the answer please :)
Chemistry
1 answer:
melamori03 [73]3 years ago
5 0
1) Calculate the volume from d = m/V => V = m/d = 2.0*10^-23 g /  1.0*10^14 g/cm^3 = 2.0*10^-9 cm^3

2) Now use the formula of volume for a sphere: V = (4/3)π(r^3) =>

r =∛[3V/(4π)] = ∛[(3*2.0*10^-9 cm^3) / (4π)] = 0.48*10^-3 cm = 4.8*10^-4 cm = 0.00048cm
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How much water must be added to 15 ml of a 2.5 M solution of sodium chloride to make a 2.0 M solution of sodium chloride?
Ivan

Answer:

3.75mL need to be added

Explanation:

M1 x V1 = M2 x V2

so V2=(M1xV1)/M2

(2.5Mx.015L)/2M= .01875L or 18.75ml

since we want to know how much needs to be added we subtract 18.75 from 15

Giving us 3.75ml need to be added

3 0
3 years ago
Identify balance chemical equation from the following
Furkat [3]

Answer:

C+O2=CO2

Explanation:

no.of atoms on reactant side= no.of atoms on product side

5 0
3 years ago
Does the law of conservation of mass apply to physical changes or just chemical changes?
slega [8]
It applies to both physical and chemical changes.
4 0
4 years ago
A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0
3 years ago
Can someone please help me with this question? Thanks in advance.
sveta [45]

Answer:

1. Fe + H2SO4 = FeSO4 + H2

iron + sulfuric acid = duretter + hydrogen

2. Mg + 2 HNO3 = Mg(NO3)2 + H2

Magnesium + nitric acid = magnesium nitrate +hydrogen

3. Zn + 2 HCl = ZnCl2 + H2

Zinc + hydrogen chloride= zinc chloride+ hydrogen

6 0
3 years ago
Read 2 more answers
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