All categories

3 years ago

How many grams of nickel (+2) are required to replace all of the silver when 15.55 grams of AgNO3 are present?

Chemistry

1 answer:

Karolina [17]3 years ago

8 0

2AgNO3 + Ni2+ = Ni(NO3)2 + 2Ag<span>+</span>

From the reaction, it can be seen that AgNO3 and Ni2+ has following amount of substance relationshep:

n(AgNO3):n(Ni)=2:1

From the relationshep we can determinate requred moles of Ni2+:

n(AgNO3)=m/M= 15.5/169.87=0.09 moles

So, n (Ni)=n(AgNO3)/2=0.045 moles

Finaly needed mass of Ni2+ is:

m(Ni2+)=nxM=0,045x58.7=2.64g

You might be interested in

Easy points!

podryga [215]

The soda an air is most likely bubbling up and is going to explode.

3 0

3 years ago

Question 14 of 25

algol13

Answer:

describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule

3 0

2 years ago

Read 2 more answers

Why are koalas so crusty?

marishachu [46]

Koalas are not crusty, but their fur is very coarse, like wool.
Hope this helps.

6 0

4 years ago

Read 2 more answers

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T $_2$ ),

P $_1$ V $_1$ / T $_1$ = P $_2$ V $_2$ / T $_2$ ,

T $_2$ = P $_2$ V $_2$ T $_1$ / P $_1$ V $_1$ ,

T $_2$ = 0.73 atm $*$ 1.8 L $*$ 298.15 K / 1 atm $*$ 1.2 L = ( 0.73 $*$ 1.8 $*$ 298.15 / 1 $*$ 1.2 ) K = 326.47425 K,

T $_2$ = 326.47425 K = 53.32425 C

7 0

3 years ago

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

3 0

3 years ago