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2 years ago

PLS HELP MEE !! Andy classified some substances as shown in the table.

Chemistry

2 answers:

Lelechka [254]2 years ago

5 0

Number 1 Bc of wood a magnetic

NISA [10]2 years ago

5 0

Answer:

The answer is leather is a non magnetic substance.

Explanation:

its kinda obvious lol

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A vitamin c tablet containing 250 mg of ascorbic acid (c6h8o6; ka=8.0×10−5) is dissolved in a 250 ml glass of water. what is the

ikadub [295]

First, we have to get the initial [C6H8O6] = mass/molar mass

when the molar mass of C6H8O6 = 176.12 g/mol

∴[C6H8O6] = 0.25 g / 176.12 g/mol
= 0.00142 M

when
C6H8O6 ⇄ H+ + C6H7O6-
intial 0.00142 M 0 0
change -X +X +X
Equ (0.00142-X) X X

so, Ka = [H+][C6H7O6-] / [C6H8O6]

by substitution:

8 x 10^-5 = X * X / (0.00142-X) by solving this equation for X

∴ X = 0.000299
∴[H+] = 0.000299

∴PH = -㏒[H+]

= -㏒ 0.000299
= 3.52

6 0

1 year ago

1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore

Alina [70]

Answer :

(1) The theoretical yield of product, $MgO$ is, 1.257 grams.

(2) The percent yield of $MgO$ is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles$

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the reaction, we conclude that

As, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 0.03142 moles of $Mg$ react to give 0.03142 moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$

$\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g$

Theoretical yield of $MgO$ = 1.257 g

Experimental yield of $MgO$ = 0.8922 g

Now we have to calculate the percent yield of $MgO$

$\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100$

$\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%$

Therefore, the percent yield of $MgO$ is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0

2 years ago

6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021

Aleks [24]

Answer:

The total amount of Au is $ $2.0\times10^{24}$

Explanation:

Given that,

Mass of 1.0 ml of Au $m=4\times10^{-2}\ g$

Total volume of water in oceans $V=1.5\times10^{21}\ L$

We need to calculate the volume in ml

Using given volume

$V=1.5\times10^{21}\times1000\ mL$

$V=1.5\times10^{24}\ mL$

We need to calculate the total mass of Au

Using given data

$1\ ml\ volume = 4\times10^{-2}\ g$

$1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}$

So, The total mass of Au is $6\times10^{22}\ g$

The mass will be in ounce,

$Mass=0.035274\times6\times10^{22}$

$Mass=2.12\times10^{21}\ ounce$

The total amount of the Au Will be

$Total\ amount=2.12\times10^{21}\times948$

$Total\ amount=2.0\times10^{24}$

Hence, The total amount of Au is $ $2.0\times10^{24}$

3 0

1 year ago

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

2 years ago

Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

Regards!

5 0

2 years ago