What type of reaction is the digestion of solid copper wire by nitric acid?

A multivitamin tablet contains 40 milligrams of potassium. how many moles of potassium does each tablet contain?

katrin2010 [14]

Conversion of mole to grams
k in mole = 1 mole/ atomic mass
K in mole =1/ 39.0983 g/mole
= 0.255765 g/mole
converting 40 grams of K
K 40 grams x [ 1 mole/ 39.0983 grams] = 1.0230623 mole
There are 1.0230623 moles of K in 40 K of Potassium

7 0

2 years ago

Read 2 more answers

Identify the following redox reactions by type. Check all that apply. (a) Fe + H2SO4 → FeSO4 + H2 combination decomposition disp

EastWind [94]

Answer :

(a) displacement reaction

(b) combination reaction

(c) disproportionation reaction

(d) displacement reaction

Explanation :

(a) The given balanced chemical reaction is,

$Fe+H_2SO_4\rightarrow FeSO_4+H_2$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Fe) replace the less reactive element (H).

(b) The given balanced chemical reaction is,

$S+3F_2\rightarrow SF_6$

This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or single product.

(c) The given balanced chemical reaction is,

$2CuCl_2\rightarrow Cu+CuCl_2$

This reaction is a disproportionation reaction in which the chemical species gets oxidized and reduced simultaneously. It is also considered as a redox reaction.

(d) The given balanced chemical reaction is,

$2Ag+PtCl_2\rightarrow 2AgCl+Pt$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Ag) replace the less reactive element (Pt).

7 0

2 years ago

This information is taken directly from Exercise 3 in the CHEM 111/112 Laboratory Manual. Another example of excellence in the p

kolbaska11 [484]

Answer:

the information was not cited correctly....

Explanation:

I hope the following explanation will help you a lot.

5 0

2 years ago

If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.

Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water = Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0

1 year ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago