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Sladkaya [172]
4 years ago
14

Object A has a length of 4cm, a width of 3cm, and a height of 1 cm. Object B is dropped into a graduated cylinder. It displaces

15 mL of water. The volume of object A is:
-greater than the volume of object b?

-less than the volume of object b?

-equal to the volume of object b?
Chemistry
1 answer:
motikmotik4 years ago
8 0

Answer:

(2) Less than the volume of object b.

Explanation:

1 mL is, by definition, 1cc (cubic centimeter) of water. Therefore if B displaces 15 cc, then its volume is 15. A's volume is 4\cdot3\cdot1 = 12cc which is less than 15, so the answer is (2).

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EQUILIBRIUM Two moles of compound P were placed in a vessel. The vessel was heated and compound P was partially decomposed to pr
Reika [66]

As given:

Initial moles of P taken = 2 mol

the products are R and Q

at equilibrium the moles of

R = x

total moles =  2 + x/2

Let us check for each reaction

A) P <-> 2Q+R

Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = 2x

moles of R = x

Total moles = (2-x) + 2x + x = 2 +2x

B) 2P <-> 2Q+R

Here x moles of P will give x moles of Q and x/2 moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = x

moles of R = x/2

Total moles = (2-x) + x + x/2 = 2 + x/2

C) 2P <-> Q+R

Here x moles of P will give x/2 moles of Q and x/2 moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = x /2

moles of R = x/2

Total moles = (2-x) + x + x = 2

D) 2P <-> Q+2R

Here x moles of P will give x/2 moles of Q and x moles of R

So at equilibrium

moles of P left = 2-x

moles of Q = x/2

moles of R = x

Total moles = (2-x) + x/2 + x = 2 + x/2

3 0
3 years ago
A 0.20 M solution of a weak acid has a pH of 5.40. What is the Ka for the acid?
sweet [91]

Answer:

The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <------> H3O+  +   A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation  (the acid form of water)

How do you calculate Ka??  Ka is the acid equilibrium constant.

( [H3O+]  . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

 HA + H2O <------> H3O+  +   A-

initial- 0.2 M             I don't have H3O+, either A-

reaction - an specific amount reacted  (X)

in equilibrium (0,2 - X)  <----->  X   +  X

And now, how's the formula for Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, soooo small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11

5 0
3 years ago
6. How does temperature relate to the layers of the atmosphere?
solniwko [45]
The atmosphere is divided into layers based on how the temperature in that layer changes with altitude, the layer's temperature gradient (Figure below). The temperature gradient of each layer is different. In some layers, temperature increases with altitude and in others it decreases.
8 0
3 years ago
A mixture of 0.307 M Cl 2 , 0.465 M F 2 , and 0.706 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g )
monitta

<u>Answer:</u> The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.307 M

Initial concentration of fluorine gas = 0.465 M

Initial concentration of ClF gas = 0.706 M

The given chemical equation follows:

                            Cl_2(g)+F_2(g)\rightleftharpoons 2ClF(g)

<u>Initial:</u>                  0.307       0.465       0.706

<u>At eqllm:</u>           0.307-x    0.465-x       0.706+2x

The expression of K_c for above equation follows:

K_c=\frac{[ClF]^2}{[Cl_2][F_2]}

We are given:

K_c=20.0

Putting values in above equation, we get:

20.0=\frac{(0.706+2x)^2}{(0.307-x)(0.465-x)}\\\\x=0.148,0.993

Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible

So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M

Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M

Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M

Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

5 0
3 years ago
Potassium hydroxide is used to precipitate each of the cations from their respective solution. determine the minimum concentrati
MAXImum [283]
This is the three cases that help to determine the minimum concentration of KOH required for precipitation 
Part a) 1.5×10^−2 M K CaCl2 
Part b) 2.3×10^−3 M Fe (NO3)2 
Part c) 2.0×10^−3 M MgBr2

a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^- 
ksp = 1.5*10^-2 + x^2 
4.68*10^-6 = 1.5*10^-2 + x^2 
x= [KOH] = 0.01766 

b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3 
Fe (OH)2 <=> Fe^2+ + 2OH^- 
ksp = 2.3*10^-3 + x^2 
4.87*10^-17 = 2.3*10^-3 + x^2 
x= 1.46*10^-7 

c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr 
Mg (OH) 2 <=> Mg^2+ + 2OH^- 
ksp = 2.0*10^-3 + x^2 
2.06*10^-13 = 2.0*10^-3 + x^2 
x= 1.015*10^-5
8 0
4 years ago
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