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2 years ago

Look at the graph

Chemistry

1 answer:

Vlad [161]2 years ago

7 0

Answer:

a item dropping

Explanation:

KE is movement and PE is height. As it's falling it gets KE and falls downward giving it more KE and less PE

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In a titration experiment, hcl and lioh solutions were used. the initial volume of hcl was 1.25ml and lioh was 2.65 ml. the fina

aliya0001 [1]

Answer is: <span>the molarity of HCl is </span>0.097 M.
Chemical reaction: LiOH + HCl → LiCl + H₂O.
V(HCl) = 13.60 mL - 1.25 mL = 12.35 mL.
V(LiOH) = 11.20 mL - 2.65 mL = 8.55 mL.
c(LiOH) = 0.140 M.
From chemical reaction: n(LiOH) : n(HCl) = 1 : 1.
c(HCl) · V(HCl) = c(LiOH) · V(LiOH).
c(HCl) = 8.55 mL · 0.140 M / 12.35 mL.
c(LiOH) = 0.097 M.

8 0

2 years ago

An atom has the following electron configuration. 1s22s22p63s23p64s? How many valence electrons does this atom have? A: 2B: 3C:

Katena32 [7]

ANSWER

2 valence electrons

STEP-BY-STEP EXPLANATION:

The electronic configuration is given below as

$1s^22s^22p^62s^23p^64s^2$

Looking at the configuration, you will see that the last energy carries 2 electrons

Hence, the valence electrons are 2

3 0

1 year ago

At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con

djyliett [7]

0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l

4 0

2 years ago

Where did coper find

ladessa [460]

Answer:

Copper is element number 29

4th row

11th column

6 0

2 years ago

A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L

slega [8]

Answer:

5.41 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

$Moles=0.033 \times {0.45}\ moles$

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01485\ g= \frac{Mass}{342.39\ g/mol}$

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer - 5.41 g</u>

5 0

3 years ago