1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
madam [21]
3 years ago
7

What are the reagents for BaCO3(s)

Chemistry
1 answer:
Wittaler [7]3 years ago
6 0

 The reagents for BaCO₃  is

    BaO and CO₂

      <em><u>Explanation</u></em>

Reagent is  a substance that bring  about a chemical reaction when added to a system.

Some  reagent  may be added to  see if a reaction has occurred.

 BaO  and  Co₂    are reagent  since they react  to produce BaCO₃ as below

BaO(s) + CO₂(g) → BaCO3(s)



You might be interested in
What are the different chemical properties of the element Cobalt?
larisa [96]

Answer:

It is a hard ferromagnetic, silver-white, lustrous, brittle element.

It is stable in air and does not react with water.

Like other metals, it can also be magnetized.

With dilute acids, it reacts slowly.

The metal melts at 1495 °C and boils at 2927 °C

Explanation:

hope this helps! maybe mark me brainliest?

8 0
3 years ago
How is the atmosphere like an ocean
miss Akunina [59]
 <span>The ocean is primarily a liquid and the atmosphere is mostly gas, so they are different states of matter. They also have different chemical compositions which really ought to be common knowledge. Also, very few fish swim in the sky.

YEP

</span>
4 0
3 years ago
Read 2 more answers
How would the pH value of an aqueous solution change, when the hydronium ion concentration is increased by a factor of 10?
Anna71 [15]

Answer:

Increasing H⁺ by 10x => pH decreases by 1 unit

Explanation:

In general, adding H⁺ ions to any aqueous solution ALWAYS causes pH values to fall ( decrease ). Just as adding OH⁻ ions to an aqueous solution causes pH values to rise ( increase ).

Here's a simple calculation demonstrating this...

Given 0.01M HCl(aq) => 0.01M H⁺(aq) + Cl⁻(aq) => pH = -log(0.01) = 2.00

Increase [H⁺] by 10x => 0.10M H⁺(aq) => pH = -log[H⁺] = -log(0.10) = 1.00

Solution with higher H⁺ concentration shows <u>pH decreasing by 1 unit.</u>

______________________________________________________-

Just to support the above statement about adding OH⁻ ions showing an increase in pH values, the following is also provided FYI ..

Given 0.01M NaOH(aq) => 0.01M OH⁻(aq) + Na⁺(aq) => pOH = -log(0.01) = 2.00 => pH = 14 - pOH = 14 - 2 = 12

Increase [OH⁻] by 10x => 0.10M OH⁻(aq) => pOH = -log[OH⁻] = -log(0.10) = 1.00 => pH = 14 - pOH = 14 - 1 = 13

Increasing [OH⁻] by 10x => <u>increasing pH by 1 unit. </u>

Solution with higher H⁺ concentration shows pH decreasing by 1 unit.

______________________________________________________

Remember, for <u>any</u> aqueous solution ...

=> Adding H⁺   => always decreases pH

=> Adding OH⁻ => always increases pH

4 0
3 years ago
Use the drop-down menus to identify the effect of each mutation.
krek1111 [17]

Answer:

1 neutral

2 beneficial

3 harmful

Explanation:

7 0
4 years ago
Read 2 more answers
PH is a logarithmic scale used to indicate the hydrogen ion concentration, [H+], of a solution: pH=−log[H+] Due to the autoioniz
ruslelena [56]

Answer:

  • A) pH = 2.42
  • B) pH = 12.00

Explanation:

<em>The dissolution of HCl is  HCl → H⁺ + Cl⁻</em>

  • To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl} = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

  • <em>The dissolution of NaOH is  NaOH → Na⁺ + OH⁻</em>
  • Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}= 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00

5 0
3 years ago
Other questions:
  • Give the formula of the conjugate base of HSO4–
    9·1 answer
  • How have chemists helped fight leukemia?
    5·2 answers
  • A sample of chemical x is found to contain 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen. The law of def
    12·1 answer
  • Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produc
    7·1 answer
  • How many atoms of oxygen is present in 10H2SO4 ?
    10·1 answer
  • To try and stimulate the correct distance, how close would you put the animal
    9·2 answers
  • What is the most abundant element on Earth?
    8·1 answer
  • How do red foxes get offspring?
    12·1 answer
  • Need answer asap
    5·2 answers
  • PLEASEEE HELPPPPP ASAPPPPPP!!!!
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!