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ArbitrLikvidat [17]
3 years ago
15

What is the mass in grams of 3.40 x 10 24 atoms he

Chemistry
1 answer:
Liula [17]3 years ago
4 0

Answer:

Avogadro's constant says that  

1

mole of any atom contains  

6.022

⋅

10

23

atoms. In this case you have  

3.40

⋅

10

22

atoms:

3.40

⋅

10

22

atoms

6.022

⋅

10

23

atoms

mol

=  

5.65

⋅

10

−

2

m

o

l

Step 2

The atomic mass of helium (He) will give you the weight of one mole of this molecule:  

1

mol =  

4.00

gram:

4.00

g

mol

⋅

5.65

⋅

10

−

2

mol

=  

0.226

g

So the  

3.40

⋅

10

22

helium atoms weigh  

0.226

gram

.

Explanation:

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damaskus [11]

Find volume of pillow

L=78cm

B=55cm

H=25cm

\\ \bull\tt\longrightarrow V=LBH

\\ \bull\tt\longrightarrow V=25(78)(55)

\\ \bull\tt\longrightarrow V=107250cm^3

\\ \bull\tt\longrightarrow V=10.72m^3

Now

Mass=5.5kg

\\ \bull\tt\longrightarrow Density=\dfrac{Mass}{Volume}

\\ \bull\tt\longrightarrow Density=\dfrac{5.5}{10.72}

\\ \bull\tt\longrightarrow Density=0.5kg/m^3

Density of water=1000kg/m^3

As it is less than density of water it will float on water

8 0
3 years ago
When are kids ready to assume adult responsibilities?
Serga [27]

Answer:

18 years old or when they become mature.

Explanation:

5 0
3 years ago
How many milliliters of 0.50 M KOH are needed
spayn [35]

Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

3 0
3 years ago
What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)
seraphim [82]

Answer:

219.15 grams

Explanation:

What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)

Mole of Na = 22.99g

Mole of Cl = 35.45g

For NaCl we have ratio of 1:1, so we have 1 Na for every Cl

So we just add the two together to get the molar mass of NaCl which is

22.99 + 35.45 = 58.44g/mol

And we know we have 3.75 moles of NaCl so we multiply that by the molar mass of NaCl to get our answer

3.75 x 58.44 = 219.15grams

6 0
3 years ago
A sample of pure NO2NO2 is heated to 335 ∘C∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2N
Alborosie

Answer:

k_c = 1. 1 × 10⁻²

Explanation:

Given that:

Temperature = 335 ° C = (335+ 273)K = 608

Pressure = 0.750 atm

Volume = 1 Litre

number of moles of NO2 = ???

Rate Constant =0.0821 L atm /K/mol

Using the Ideal gas equation

PV = nRT

n = \frac{PV}{RT}

n = \frac{0.75*1}{0.0821*608}

n = 0.015

n = 1.5 × 10⁻² mole

Density = 0.525 g/L

The equation for the reaction can be illustrated as:

                     2NO2(g)         ⇌          2NO(g)         +         O2(g)

For the ICE table; we have:

 

Initial                 x                                   0                              0

Change            -2y                               + 2y                          +y

Equilibrium        (x - 2y)                        2y                             y

Total moles at equilibrium = (x-2y)+2y+y

= x + y moles

However,

1.5 × 10⁻² mole of the mixture has a mass of 0.525 g

i.e x + y moles = 1.5 × 10⁻² mole

Now, molar mass of 1 mole of NO2 = 46g/mol

Since number of moles = \frac{mass}{molar mass}

mass of (x-2y) moles = 46 × (x-2y) g

Molar mass of NO = 30 g/mol

Also, mass of NO = 2y × 30 = 60y

Molar mass of O2 = 32 g/mol

Mass of O2 = y × 32 = 32y

Total mass = ( 46x - 90y)+60y+32y = 0.525

46x = 0.525

x = \frac{0.525}{46}

x = 0.0114

x = 1.14 × 10⁻²

x + y moles = 1.5 × 10⁻²

y =  1.5 × 10⁻² -  1.14 × 10⁻²

y = 0.0036

y = 3.6 × 10⁻³

At equilibrium

[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M

[NO] = 2 ( 3.6 × 10⁻³)  = 7.2 × 10⁻³ M

[O2] = 3.6 × 10⁻³ M

k_c = \frac{[NO]^2[O_2]}{[NO_2]^2}

k_c = \frac{(7.2*10^{-3})^2(3.6*10^{-3})}{(4.2*10^-3)^2}

k_c = 0.011

k_c = 1. 1 × 10⁻²

4 0
3 years ago
Read 2 more answers
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