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nikitadnepr [17]
3 years ago
14

How many moles contain 1.25x 10^15 molecules carbon dioxide?

Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0
Avagadros number states that 1 mol of any substance is made of 6.022 x 10²³ units.
These units could be atoms that make up an element or molecules that make up a compound. CO₂ is a compound therefore its made of CO₂ molecules.
1 mol of CO₂ contains 6.022 x 10²³ molecules of CO₂
Therefore if  6.022 x 10²³ molecules of CO₂ make up 1 mol 
then 1.25 x 10¹⁵ molecules of CO₂ make up - \frac{1.25 x 10¹⁵}{6.022 x 10²³}
Number of CO₂ moles present - 2.0756 x 10⁻⁹ mol of CO₂
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2CuS + 3O2 = 2CuO + 2SO2
Nata [24]
I have no yuuuuu to say anything abt to
4 0
3 years ago
The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0
3 years ago
Most reactive<br> O Be<br> O Mg<br> O Ba<br> o Ca<br> Which one is the most reactive
katen-ka-za [31]

Answer:

Mg

Explanation:

I think the answer is Mg

7 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
Consider the reaction below. HI + H2O mc014-1.jpg H3O+ + I– Which is an acid-conjugate base pair? HI and H2O H2O and H3O+ H3O+ a
lesya692 [45]

Answer: Option (d) is the correct answer.

Explanation:

According to Bronsted-Lowry, species which donate a proton are known as acid. The species which accept a proton are known as a base.

In the given reaction, acids and bases are as follows.

    HI  +   H_{2}O \rightarrow    H_{3}O^{+}           +         I^{-}

 Acid        Base     Conjugate acid   Conjugate base

Therefore, the acid HI loses a proton to form a conjugate base that is I^{-}.

Thus, we can conclude that HI and I^{-} is an acid conjugate base pair.

4 0
3 years ago
Read 2 more answers
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