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3 years ago

Sand and salt are poured into a single beaker. the result is a mixture because

1 answer:

6 0

Sand and salt are poured into a single beaker. the result is a mixture because they are not chemically combined. When water is added to the sand-salt mixture, salt dissolves in water, and sand does not. When the liquid is poured through a filter, salt water passes through the filter. The remaining salt water is heated until water is boiled off (evaporated) leaving salt behind.

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Government should use informed science to help make policies protect all citizens<br> True or False

Sloan [31]

Answer:

yes

Explanation:

8 0

3 years ago

An -ate or -ite at the end of a compound name usually indicates that the compound contains _____.

liberstina [14]

A polyatomic ion.

Examples: Sulfate, Sulfite, Nitrate, Nitrite...

6 0

3 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

6 0

3 years ago

If a chemist has a stock solution of HBr that is 10.0 M and would like to make 450.0 mL of 3.0 M HBr, how would he do it?

asambeis [7]

This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution

Following dilution formula can be used

c1v1 = c2v2

Where c1 is concentration and v1 is the volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting these values

10.0 M x v1 = 3.0 x 450.0 mL

v1 = 135.0 mL

A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M

4 0

3 years ago

Guys, please, please help!!

shutvik [7]

Ookay

1) T
2) T
3) F

Hope i helped :)

6 0

3 years ago

Read 2 more answers