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Archy [21]
2 years ago
13

Help, please!! If an object with a mass of 4.2 kg accelerates at 2.6 m/s2, what was the force exerted?

Chemistry
1 answer:
Paha777 [63]2 years ago
6 0
10.92N

Force = mass x acceleration
4.2kg x 1.6m/s^2 = 10.92N
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What mass of oxygen is needed to burn 4g of hydrogen, given the equation below
aleksandrvk [35]

32g of oxygen is required to burn 4g of hydrogen.

Define molecular mass.

A specific molecule's mass is expressed in daltons and is known as the molecular mass (m) (Da or u). Due to the varying isotopes of an element that they contain, multiple molecules of the same substance can have distinct molecular weights.

The total atomic mass of every atom in a molecule, calculated using a scale with hydrogen, carbon, nitrogen, and oxygen having atomic masses of 1, 12, 14, and 16, respectively. For instance, water has a molecular mass of 18 (2 + 16), which consists of two hydrogen atoms and one oxygen atom. known also as molecular weight.

In ,2H2+O2-----> 2H2O

H 2 molecules have a mass of 2 g/mol.

The molecular weight of oxygen is 32 g/mol.

When the chemical equation is balanced,

To totally react, 32 g of oxygen are needed for every 22=4 g of hydrogen.

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7 0
1 year ago
I WILL GIVE BRAINLIEST!!!!!
Alexxx [7]
                                    HNO₃ + H₂S → S + NO + H₂<span>O

Assign Oxidation Number:

                                L.H.S                               R.H.S
           N in HNO</span>₃  =  +5                                     +2  =  N in NO
               S in H₂S  =  -2                                       0  =  S in S

Write Half cell Reactions:
Reduction Reaction:
                                 
3e⁻  +   HNO₃   →   NO      -------(1)
Oxidation Reaction:
                                  H₂S   →   S  +  2e⁻             -------(2)

Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons. 

                                          6e⁻  +   2 HNO₃   →   2 NO
                                    
                                             3 H₂S   →   3 S  +  6e⁻
                                                                                  Cancel e⁻s,
                                 ______________________________
                               
                             2 HNO₃  +  3 H₂S   →   2 NO  +  3 S  +  H₂O

Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.

                            2 HNO₃  +  3 H₂S   →   2 NO  +  3 S  +  4H₂O


                                
4 0
3 years ago
You would be most likely to use a slicing machine if you were using the __________ method to produce cookies.
Verizon [17]

You would be most likely to use a slicing machine if you were using the <u>icebox </u>method to produce cookies.

In the icebox method a type of cookie in which the dough is made, rolled into a stick, and refrigerated until the dough hardens. The dough can be removed from the refrigerator, cut into individual pieces, and then baked. The rest of the dough is returned to the refrigerator until needed.

Icebox method, also known as refrigerator cookies, are sliced ​​and baked cookies. The dough is formed into logs, chilled in the refrigerator (also called an icebox), sliced ​​, and then baked.

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7 0
1 year ago
Relation between glancing and angle of deviation<br>​
shtirl [24]

Answer:

the Glancing angle is the angle between the incident ray and plane mirror which is 90o in the given case. The angle between the direction of the incident ray and the reflected ray is the angle of deviation. Since the angle of deviation for a plane mirror is twice the glancing angle, the angle of deviation is 1800.

5 0
2 years ago
One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w
4vir4ik [10]

Answer:

4

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}

Z fro Be = 4

39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0
3 years ago
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