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3 years ago

Can someone find the independent variable and the dependent variable ? :)))

Chemistry

1 answer:

sasho [114]3 years ago

7 0

Answer: Dependant variable = The distance it takes for each car to stop. Independent variable = Both cars go at 60 mph.

Explanation:

This is right because the dependant variable changes and the independent variable is the outcome of the dependant variable.

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What is the pOH of 0.50 molar H3BO3?

Crazy boy [7]

Answer:

A. 10.25

Explanation:

Pkb =4.77

So pka = 14 - pka = 9.23

$Ka =10^{-pka}$

$H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)$

Initial 0.50M 0 0

Change -x +x +x

Equilibrium 0.50M-x +x +x

$Ka =\frac {((x)(x))}{(0.50M-x)}$

$5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}$

(-x is neglected) so we get

$5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}$

$x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}$

$pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76$

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

4 0

3 years ago

Is it wierd if i think my step sis is thick

iren2701 [21]

Answer:

yes it is and sometimes it's not

8 0

2 years ago

Read 2 more answers

Decrire l ebullition de l eau

meriva

L'eau se réchauffe elle s'intesifie et commnce a bouilloner ensuite L'eau s'évapore

7 0

3 years ago

Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.

Illusion [34]

Answer:

- 130.64°C.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

∴ T₂ = V₂T₁/V₁ = (307.0 L)(294.0 K)/(634.0 L) = 142.36 K.

∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.

5 0

3 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago