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3 years ago

What is the formula for tetrasulfur hexabromideIf you answer I will give you a brainliest

Chemistry

1 answer:

gavmur [86]3 years ago

6 0

Answer:

Tetrasulfur Dinitride S4N2

Tetrasulfur pentoxide S4N4

Tetrasulfur tetranitride N4S4

Tetrasulfur pentoxide S4N4

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Which place in the world has the highest average annual precipitation?

alekssr [168]

Answer: Colombia

Explanation:

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2 years ago

Read 2 more answers

How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

7 0

3 years ago

How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid wat

lozanna [386]

The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is 118.16 cal

calculation

Heat in calories = MCΔ T where,

M(mass)= 2.18 g
C(specific heat capacity)= 1.00 cal/g/c
ΔT( change in temperature)= 69.5- 15.3 =54.2 c

heat is therefore= 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal

7 0

3 years ago

Read 2 more answers

MULTIPLE CHOICE

Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0

3 years ago

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

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

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

2 years ago