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vitfil [10]
3 years ago
12

What does airbag, used for safety of car driver, contain?

Chemistry
2 answers:
sammy [17]3 years ago
7 0
B is the correct answer for sure.)
MrRa [10]3 years ago
6 0
<span>Answer : B. hope it helped</span>
You might be interested in
Will make brainliest
Dafna11 [192]

Answer:

Boron

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

For example

In given atomic model we can see that there are five electrons out side the nucleus it means this atomic model is of that element which have atomic number 5 and it is boron.

It has 5 electron 5 proton and 6 neutrons.

It means its atomic number is five and mass number is 5+6 = 11

8 0
3 years ago
What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution
KATRIN_1 [288]

Answer:

0.8M

Explanation:

CM=n/V

8 0
3 years ago
How many atoms are there in 4.13 moles of nickel (Ni)
PtichkaEL [24]

Answer:

24.87× 10²³ atoms of Ni

Explanation:

Given data:

Number of atoms of Ni= ?

Number of moles of Ni = 4.13 mol

Solution:

we will calculate the number of atoms of Ni by using  Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

4.13 mol ×  6.022 × 10²³ atoms /1 mol

24.87× 10²³ atoms of Ni

6 0
3 years ago
Write the balanced chemical equation for the reaction between aqueous sodium carbonate and
ioda

Answer:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

Explanation:

This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:

HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)

However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

7 0
3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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