All categories

3 years ago

What is the proper value for D in the above table. A. 65 B. 35 C. 30 D. 63

Chemistry

2 answers:

tankabanditka [31]3 years ago

8 0

It’s A or C but I think A so just go with A

Zolol [24]3 years ago

5 0

Answer:

Explanation:

Atomic Number, Protons, and Neutrons are always of the same amount.

You might be interested in

Why is seawater a better conductor of electricity than freshwater?

GenaCL600 [577]

Answer:

please refer to the attachment above;)

3 0

3 years ago

Read 2 more answers

78.5 mol of P4O10 contains how many moles of P?

MArishka [77]

4 x 36 moles. of Phosporus and 10 x 36 of Oxygen. I hope this helps. (:

7 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

Please help thank you (15 points)

Eva8 [605]

Answer:

D. The sun has less gravitational pull on Pluto because it is farther away.

6 0

3 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago