All categories

3 years ago

BRAINLIEST IF RIGHT!! Write 200.0398 in 2 sig figs.

Chemistry

2 answers:

IgorC [24]3 years ago

8 0

Answer:

step by step explanation:

Archy [21]3 years ago

3 0

The answer would be 200

You might be interested in

Help please!!!!!!!!!!!!!!!!

Romashka-Z-Leto [24]

Answer:

A....................

3 0

3 years ago

0.023 M solution of perchloric acid

kogti [31]

The original question is to find the pH and the pOH of 0.023 M of perchloric acid.

Answer:
pH = 1.638
pOH = 12.362

Explanation:
1- getting the pH:
pH can be calculated using the following rule:
pH = -log[H+]
Since the given solution is an acid, this means that [H+] is the same as the concentration of the solution.
This means that:
[H+] = 0.023
Substitute in the above equation to get the pH as follows:
pH = -log[0.023]
pH = 1.638

2- getting the pOH:
We know that:
pH + pOH = 14
We have calculated that pH = 1.638.
Substitute in the above equation to get the pOH as follows:
pOH + 1.638 = 14
pOH = 14 - 1.638
pOH = 12.362

Hope this helps :)

5 0

3 years ago

1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant

vodomira [7]

The masses of the components are obtained as;

Sodium hydrogen carbonate = 3.51 g
Sodium carbonate = 8.708 g

<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

$2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)$

Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g

Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol

= 3.51 g

Learn more about anhydrous sodium carbonate :brainly.com/question/20479996

#SPJ1

6 0

1 year ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

3 years ago

What is a real life example of Charles’ law?

max2010maxim [7]

Leaving a basketball out in the cold weather. When a basketball if left in a cold garage or outside during the cold months, it loses its air inside (or volume).

6 0

3 years ago