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3 years ago

What happens to halogens on going top to bottom in their group

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

3 0

they remain as they are and don't change their type

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Please Answer ASAP!!

Roman55 [17]

Answer:

Explanation:

6 0

3 years ago

NO<br> Assign oxidation numbers to each element in this compound.

marta [7]

Answer:

N=+2

O=-2

Explanation:

The compound NO is electrically neutral.

Lets assign the oxidation number of nitrogen to be N. The oxidation number of oxygen (-2) is then used as a reference.

For the compound to have a zero charge, sum of the oxidation numbers equals zero.

N+ (-2)=0

N=+2

O=-2

7 0

4 years ago

Calculate the molarity of the two solutions.

beks73 [17]

Answer:

A = Molarity = 0.22 M

B = Molarity = 0.36 M

Explanation:

Given data:

For first solution:

number of moles = 0.550 mol

Volume of solution = 2.50 L

Molarity = ?

Molarity:

Formula:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.550 mol / 2.50 L

Molarity = 0.22 M

For second solution:

Mass of NaCl = 15.7 g

Volume of solution = 709 mL or 709/1000 = 0.709 L

Molarity = ?

Solution:

Number of moles = mass / molar mass

Number of moles = 14.7 g/ 58.44 g/mol

Number of moles = 0.252 mol

Molarity:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.252 mol / 0.709 L

Molarity = 0.36 M

8 0

3 years ago

What is the chemical formula for toothpaste?

drek231 [11]

The most common type of fluoride found in toothpaste is part of the compound sodium fluoride which has the chemical formula NaF

4 0

4 years ago

Read 2 more answers

Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....

Oduvanchick [21]

Answer:

Explanation:

$\mathbf{From \ the \ information \ given:} \\ \\ \mathbf{The \ rates \ of \ the \ f ollowing \ reactions \ can \ be \ expressed \ as \ follows:}$

(a)

$\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}$

(b)

$\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}$

(c)

$\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }$

(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

$\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}$

5 0

3 years ago