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3 years ago

What do scientists use to increase the surface area of a solute? heat a beaker a mortar and pestle a flask a stir bar

Chemistry

2 answers:

zhenek [66]3 years ago

6 0

A motor and pestle C

Novay_Z [31]3 years ago

4 0

Mortar and pestle could be used to break solute into smaller pieces, which would increase surface area

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A given mass of oxygen occupies 500 ml when the

matrenka [14]

Ok we can use boyle’s law (stating that P is proportional to V) to make the equation (P1V1) =(P2V2).
once we’ve done this, we can plug in the numbers:
(800•500) = (200•V2)
and then we get that
V2= 2000 ml

hope this helps!! :)

5 0

3 years ago

A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t

tresset_1 [31]

Answer: The percent gallium in gallium bromide is 30.30 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

$\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol$

The chemical equation for the reaction of gallium bromide and silver nitrate follows:

$GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2$

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = $\frac{1}{1}\times 0.00072=0.00072moles$ of gallium nitrate

Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

$0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g$

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = $\frac{69.72}{193.73}\times 0.139=g$

To calculate the percentage composition of gallium in gallium bromide, we use the equation:

$\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100$

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

$\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%$

Hence, the percent gallium in gallium bromide is 30.30 %.

3 0

3 years ago

Explain why the elements of groups 1 and 7 are mostly used in the form of compound

docker41 [41]

Group 1 and group 7 of element group consists of alkali metal, which are generally combined with other elements in nature, so they have properties of compound

8 0

3 years ago

Determine the LIMITING reactant in the following balanced equation:

padilas [110]

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂ → 2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

KBr : KCl

2 : 2

0.03 : 0.03

KBr : Br₂

2 : 1

0.03 : 1/2×0.03= 0.015

Cl₂ : KCl

1 : 2

0.09 : 2/1×0.09 = 0.18

Cl₂ : Br₂

1 : 1

0.09 : 0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.

5 0

3 years ago

How to find the charge of an element

katrin2010 [14]

Answer:

the atomic number is how you find the charge

7 0

3 years ago