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2 years ago

How many kilograms are in 4 moles of Na2CO3?

1 answer:

6 0

First you need to find the amount of mass of Na2CO3 in one moles
(Use periodic chart)

Na= 22.99 x 2 = 45.98
C = 12.01
O = 16.00 x 4 = 64.00

Add the molar masses together to get 121.99

To find how many grams are in 4 moles, times 121.99 by 4
This gives you 487.96

But the questions asks for the answer to be in kilograms nor grams, to change into kilograms divide by 1000
This gets you the answer: 0.49 kg

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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

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

Sergio039 [100]

So 100 g of this substance has 63.57 g of carbon, 6 g of hydrogen, 9.267 of nitrogen, and 21.17 of oxygen. I need to have them all in moles (n). You can find the molar mass (M) of each element in a periodic table.

n = m/M
63.57 g C -> 63.57 g C/12.01 g/mol = 5.29 moles C
6 g H -> 6 g C/1.008 g/mol = 5.95 moles H
9.267 g N -> 9.267 g N/14.01 g/mol = 0.6615 moles N
21.17 g O -> 21.17/16.00 g/mol = 1.32 moles O

So the minimum formula has this rate:

C 5.29 H 5.95 N 0.6615 O 1.32

Now you should divide all those numbers by the smallest one (1.32):

C 4 H 4.5 N 0.5 0 1

Now it looks a lot more like a molecular formula… but we still have fractions.

Let’s multiply all numbers by 2:

C8H9N1O2

<span>Now they are all whole numbers!
</span>

So the minimum formula is C8H9NO2

The minimum formula is not always equal to the molecular formula… but in this case, I found there is a molecular formula with this same numbers, and it’s called acetaminophen.

5 0

2 years ago

Read 2 more answers

Unit Conversion Help Thank you

AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

$\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}$

$=1721.72\text{ g/qt}$

Therefore, 1721.72 g/qt are in 18.2 g/cL

5 0

2 years ago

HELP

agasfer [191]

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02 mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02 mol × 6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

5 0

1 year ago

Consider the equilibrium reaction. 4A+B↽−−⇀3C After multiplying the reaction by a factor of 2, what is the new equilibrium equat

disa [49]

8A+2B——> 6C
since you multiply by a factor of 2 you do that to each letter
4*2=8
1*2=2
3*2=6

8 0

2 years ago