Whats the answer of dependent variables and independent varies

The formula for a compound of Li+ ions and Br- ions is written LiBr. Why can’t it be written Li2Br? Why isn’t it written BrLi?

maksim [4K]

Lithium has charge of +1 and bromide has charge of - 1. So they combine to form the compound lithium bromide which is expressed as LiBr.

<h3>Explanation:</h3>

Lithium is an alkali metal placed in group 1 or periodic table. It has a valency of 1 which is achieved as lithium loses an electron to achieve a charge of +1.

Bromine is a halogen which is placed in group 17 of periodic table. It has a valency of 1 which is achieved as bromine looses an election to achieve a charge of - 1.

Lithium is the cation and bromide is the anion. So lithium is written in front and bromine following the cation. And as both of their valencies are 1, so they form the compound LiBr.

3 0

3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

1. Table

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

2. Orders

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

3. Rate equation

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

4. Rate constant, k

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

5. Rate law:

$rate=k[A]^2C=0.0075[A]^2[C]$

6. Substitute

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

Which is the correct measure of density?

Goryan [66]

$\huge \sf༆ Answer ༄$

Density is defined as Mass per unit volume of an object ~

And hence, the correct choice is ~ 1

$\sf \dfrac{mass}{volume}$

$꧁ \: \large \cal{TeeN \: ForeveR } \: ꧂$

6 0

2 years ago

Read 2 more answers

Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium

pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

$k_{c}=3.07\times10^{-4}$

Temperature = 24°C

We know that,

The balance equation is

$2NOBr\Rightarrow 2NO+Br_{2}$

Initial concentration is,

$0.878\Rightarrow 0+0$

Concentration is,

$-2x\Rightarrow 2x+x$

Equilibrium concentration

$0.878-2x\Rightarrow 2x+x$

We need to calculate the value of x

Using formula of concentration

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

$2x^2=0.0002367+0.001228x^2-0.0010536x$

$2x^2-0.001228x^2+0.0010536x-0.0002367=0$

$1.998772x^2+0.0010536x-0.0002367=0$

$x=0, 0.01062$

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

$concentration\ of\ NO=2\times0.01062$

$concentration\ of\ NO=0.02124$

Hence, The equilibrium concentration of NO is 0.02124 M.

7 0

4 years ago

What is true of a basic solution at room temperature? it has a ph value below 7. it has a greater concentration of hydroxide com

slega [8]

The true statement about basic solution at room temperature is that it has a greater concentration of hydroxide compared to hydronium ions.
Basic solutions have always pH greater than 7.
Basic solutions have bitter and caustic taste.
Basic solutions are not used as conductors in car batteries, acidic electrolytes are used in car batteries.

7 0

3 years ago