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Alinara [238K]
3 years ago
7

How many moles of mgci2 are there in 332 g of the compound

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0
Find molar mass of MgCl2 so M=<span>95.211 g/mol
N=m/M 
N=332/95.2
N=3.487
dont forget to round your sig figs </span>
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10 examples of<br>combination reaction​
Alla [95]

You can easily find hundreds on the internet.

4 0
3 years ago
Read 2 more answers
Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
Flauer [41]
I’m assuming you mean barium nitrite, Ba(NO2)2.

First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.

45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
4 0
2 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
Boron<br> B<br> 10.811<br> Number of neutrons
andrew-mc [135]

Answer:

6

Explanation:

To determine the number of neutrons we round 10.8 to 11 and subtract the atomic number (5) and get 6; therefore, boron has 6 neutrons.

7 0
3 years ago
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