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2 years ago

In balancing an equation, we change the __________ to make the number of atoms on each side of the equation balance.

Chemistry

1 answer:

Gnesinka [82]2 years ago

7 0

If you are talking about chemistry, it's coefficient!

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What makes an acid a strong acid in water makes a base or weak base

11Alexandr11 [23.1K]

It’s ph level on the ph scale. 1-3 is a strong acid, 4-6 is a weak acid, 7 is neutral, 8-10 is a weak base, and 10-14 is a strong base.

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2 years ago

In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

gladu [14]

Answer:

$\large \boxed{\text{183 g}}$

Explanation:

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}$

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2 years ago

Students are measuring the length of time it takes for soda to go flat at three different temperatures. Which type of graph shou

Akimi4 [234]

The answer would be a bar graph

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3 years ago

When a substance is in the process of changing phases states of matter what happens to the temperature

Reptile [31]

Answer:

Temperature stays constant.

Explanation:

During a phase change, the temperature does not change. Temperature will only change when a phase change is completed.

8 0

2 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

3 years ago