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Hatshy [7]
2 years ago
15

You may expect _____

Chemistry
1 answer:
TiliK225 [7]2 years ago
5 0

Answer:

covalent..............

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Ok, I need help with this one!!!!!!!!!!!!!!!!!!!! plz and thank you
Kitty [74]

Answer:

D)

Explanation:

3 0
3 years ago
A ball is rolled along and off a table. Which of the following acts on the ball after it leaves the
alexira [117]
Both the force of the earth’s gravity on the ball and the force the ball got from being rolled off
6 0
3 years ago
A sample of gas is held at 1000C at a volume of 20 L. If the volume is increased to 40 L, what is the new temperature of the gas
kaheart [24]

Answer:

The new temperature will be 2546 K or 2273 °C

Explanation:

Step 1: Data given

The initial temperature = 1000 °C =1273 K

The volume = 20L

The volume increases to 40 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 20L

⇒with T1 = the initial temperature = 1273 K

⇒with V2 = the increased volume = 40L

⇒with T2 = the new temperature = TO BE DETERMINED

20L/ 1273 K = 40L / T2

T2 = 40L / (20L/1273K)

T2 = 2546 K

The new temperature will be 2546 K

This is 2546-273 = 2273 °C

Since the volume is doubled, the temperature is doubled as well

8 0
3 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

Now we have to calculate the moles of Fe_2O_3

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg

Now we have to calculate the percent purity of Fe_2O_3 in the original sample.

Mass of original sample = 2.86\times 10^3kg

\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100

\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

Therefore, the percent purity of Fe_2O_3 in the original sample is 87.94 %

3 0
3 years ago
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
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