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ch4aika [34]
4 years ago
11

Where are chemicals found in the home?

Chemistry
2 answers:
Thepotemich [5.8K]4 years ago
8 0
Chemicals are find in C. In every room
vodomira [7]4 years ago
6 0

Answer:

c

Explanation:

chemicals can be found in every part of our lives

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Which of the following is an example of the geosphere interacting with the cryosphere? (5 points)
In-s [12.5K]

erosion is geosphere and glaciers are cyrosphere

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3 years ago
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What is the angle between the carbon-sulfur bond and the carbon-nitrogen bond in the thiocyanate ( SCN− ) ion?
dexar [7]
I’m not completely sure
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3 years ago
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Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.
Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

HCOOH\leftrightarrow H^++HCOO^-

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}

1.8*10^-^4=\frac{x^2}{c-x}

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

1.8*10^-^4=\frac{x^2}{0.12}

Let's solve this for x. Multiply both sides by 0.12

2.16*10^-^5=x^2

taking square root to both sides:

x=0.00465

Now, we have got the concentration of [H^+] .

[H^+] = 0.00465 M

We know that, pH=-log[H^+]

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

7 0
3 years ago
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Can someone plz help me? :(
Maslowich

Answer:

Its C and D I believe

Explanation:

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3 years ago
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1. Metallic strontium crystallizes in a face-centered cubic lattice, with one Sr atom per lattice point. If the edge length of t
Zarrin [17]

Answer:

r=215pm

N_{Mn}=20

Explanation:

From the question we are told that:

Edge length of the unit cell l=608pm

a)

Generally the equation for The relationship between edge length and radius is mathematically given by

4r=\sqrt{2a}

Therefore

4r=\sqrt{2*608}

r=\frac{\sqrt{2*608}}{4}

r=215pm

b)

From the question we are told that:

Density \rho=7.297

Edge length of l=630.0 pm=>630*10^-{10}

Therefore Volume  is given as

V=l^3

V=630*10^-{10}^3

V=2.50047*10^{−22}

Generally the equation for Mass is mathematically given by

m=Volume*density

m=V*\rho

m=2.50047*10^{−22}*7.297

m=1.83*10^{-21}g

Therefore Molarity is given as

n=\frac{M}{Molar M}

n=\frac{1.83*10^{-21}g}{55}

n=3.32*10^{-23}

Finally The atoms in a unit cell is

N_{Mn}=Moles*Avogadro\ constant

N_{Mn}=3.32*10^{-23}*6.023*10^{23}

N_{Mn}=20

7 0
3 years ago
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