Convert 345 g to kg Determine the conversion factor and the answer (1,000 g = 1 kg)

First to answer correctly gets brainleist

Brums [2.3K]

When it is summer in the Northern Hemisphere, it is winter in the Southern Hemisphere. It could be A I’m not sure if this helps.

6 0

3 years ago

43 Heat is the most common form of energy absorbed or released in chemical reactions. TRUE FALSE

mariarad [96]

Yup. True. I guess the next would be light energy

5 0

4 years ago

Read 2 more answers

A 30.00 mL sample of HBr was titrated using 0.1756 M NaOH. Determine the concentration of the HBr sample if 17.52 mL of NaOH was

postnew [5]

Answer: The concentration of HBr sample is 0.1025 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HBr

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=30.00mL\\n_2=1\\M_2=0.1756M\\V_2=17.52mL$

Putting values in above equation, we get:

$1\times M_1\times 30.00=1\times 0.1756\times 17.52\\\\M_1=\frac{1\times 0.1756\times 17.52}{1\times 30.00}=0.1025M$

Hence, the concentration of HBr sample is 0.1025 M

5 0

3 years ago

Hich one of the following statements about orbital hybridization is incorrect?

photoshop1234 [79]

Answer : The incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(a) The given molecule is, $CH_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(b) The given molecule is, $CO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(c) The given molecule is, $NH_3$

$\text{Number of electrons}=\frac{1}{2}\times [5+3]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 3 atoms around the central nitrogen atom, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.

(d) sp² hybrid orbitals are coplanar, and at 120° to each other.

The sp² hybrid orbitals has trigonal planar geometry and are in the same plane that means coplanar. Thus, the bond angle is 120° to each other.

(e) sp hybrid orbitals lie at 180° to each other.

The sp hybrid orbitals has linear geometry. Thus, the bond angle is 180° to each other.

Hence, the incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.

4 0

3 years ago

2 2NH3 N, + 3H. O 9 mol H, X 2 = C

MaRussiya [10]

Ummm...

WHat is this? What do you want us to do??

4 0

3 years ago