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3 years ago

What’s the answer????

Chemistry

2 answers:

ivolga24 [154]3 years ago

6 0

Answer:

DNA

Explanation:

adoni [48]3 years ago

5 0

I think dna not sure tho only answering so I could ask questions

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Help? for me? yes? please?

Sladkaya [172]

Answer:

2 Mg + O2 -> 2 MgO

is a synthesis reaction.

6 0

2 years ago

Write the balanced equation for the neutralization reaction between HCl and Ba(OH)2 in an aqueous solution. Phases are optional.

ad-work [718]

Explanation:

2HCl + Ba(OH)2 ------> BaCl2 + 2H2O

4 0

2 years ago

WILL MARK BRAINLIEST!!!!

Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Compounds
Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

Step 2: Identify Conversions

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})$
[DA] Divide [Cancel out units]: $\displaystyle 5.7015 \ mol \ K_2SO_4$

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0

3 years ago

In a lab, a student dissolves 6.9 g of sodium chloride (NaCl) in 125 g of water (H2O).

kati45 [8]

Hope it cleared your doubt.

5 0

2 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

2 years ago