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2 years ago

How many moles of H2SO4 are needed to prepare 200 mL of a 1.0 M solution?

Chemistry

1 answer:

jok3333 [9.3K]2 years ago

4 0

The answer is 0.2moles

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How often do most coastlines on Earth experience two high tides and two low tides?

Elza [17]

Answer:

24 hours and 50 minutes

Explanation:

Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience two high and two low tides every 24 hours and 50 minutes. High tides occur 12 hours and 25 minutes apart. It takes six hours and 12.5 minutes for the water at the shore to go from high to low, or from low to high.

4 0

2 years ago

11. Is this compound an acid or a base: H20?<br> both<br> base<br> neutral<br> acid

dolphi86 [110]

Answer:

Neutral

Explanation:

It has a pH of 7, so it is considered neutral, and neither an acid nor a base. This means it is a neutral!

6 0

2 years ago

Read 2 more answers

CsBr formula name???

olganol [36]

Answer

PubChem CID/molecular formula

Explanation:

Cesium bromide

PubChem CID 24592

Molecular Formula CsBr or BrCs

Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...

Molecular Weight 212.81 g/mol

Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)

have a good day /night

may i please have a branllist

4 0

2 years ago

A salt shaker filled with 17.0g sodium chloride contain how many moles?

kati45 [8]

The molecular formula for sodium chloride is NaCl. The sum of their atomic weights is (22.99 grams/mole + 35.45 grams/mole) = 58.44 grams/mole
take (17.0 grams)/(58.44 grams/mole), which equals 0.291 moles of NaCl.

8 0

3 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

6 0

6 months ago