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3 years ago

Metals tend to __ electrons and form _ ions.

Chemistry

1 answer:

masya89 [10]3 years ago

5 0

Metals tend to lose electrons and form electro-positive ions / cations.

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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

What are easy ways to Study Chemistry ??

Elina [12.6K]

Answer:

seneca

past papers

take notes from videos

Explanation:

very good website, asks questions about the subject correct for your exam board and gives correct answers and explanations

exam papers always help

7 0

2 years ago

What would the products be for the reaction between Na3PO4 + MgSO4?

ivann1987 [24]

MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2

Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2

Explanation:

7 0

2 years ago

2. What types of organisms have cells with very large vacuoles?

Sedaia [141]

The answer would be plant cells

5 0

3 years ago

Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c

jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, $Mg$ = 0.100 mol

Moles of hydrochloric acid, $HCl$ = 0.500 mol

According to the reaction shown below:-

$Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}$

1 mole of Mg reacts with 2 moles of $HCl$

0.100 mol of Mg reacts with 2*0.100 mol of $HCl$

Moles of $HCl$ must react = 0.200 mol

Available moles of $HCl$ = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus, $Mg$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $Mg$ on reaction forms 1 mole of $H_2$

0.100 mole of $Mg$ on reaction forms 0.100 mole of $H_2$

Mole of $H_2$ = 0.100 mol

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 2.24 L

2.24 L of hydrogen gas, measured at STP, are produced.

4 0

3 years ago

Metals tend to ________ electrons and form _______ ions.

Metals tend to __ electrons and form _ ions.