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3 years ago

How many grams are present in 3.4 moles of CH2COOH? Show all work

Chemistry

1 answer:

joja [24]3 years ago

3 0

Answer:

m = 200.6 gr

Explanation:

CH2COOH : n = 3.4 mol

Mr = 12 + 2 + 12 + 2×16 + 1 = 59

n= m / Mr => m = n× Mr => m = 3.4×59 => m = 200.6 gr

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Mars is known as the ____ planet.<br><br> A) Blue<br> B) Gaseous<br> C) Red<br> D) Sandy

Leni [432]

Answer:

Of course it's C

Red planet

Explanation:

It is because the soil on Mars is rich of Fe (Iron).

That makes the soil look red.

Even on our planet we have such this places like hormuz island in Iran.

5 0

3 years ago

For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

Now, we conclude that the oxidation number increases this means $KNO_{3}$ itself get oxidized to $O_{2}$ and it can act as reducing agent.

4 0

4 years ago

Read 2 more answers

What volume of H2 at STP, can be released by 42.7 g Zn in the equation: Zn + 2 HCl ————> H2 + ZnCl2

ehidna [41]

Answer:

0.86L.

Explanation:

1mol of Zn has mass of 65.39g . The amount of Zn is 2.5g65.39g/mol=0.038mol

7 0

2 years ago

How many moles are in 5.6x1025 molecules of calcium oxide?

Vlad1618 [11]

Answer:

<h2>93.02 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{5.6 \times {10}^{25} }{6.02 \times {10}^{23} } \\ = 93.023...$

We have the final answer as

<h3>93.02 moles</h3>

Hope this helps you

8 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

Allisa [31]

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water = Heat gain by iron rod

$m_{w}$ $C_{w}$ ( $T_{2}$ - $T_{f}$ ) = $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm

8 0

3 years ago