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Kipish [7]
3 years ago
13

AYUDA DOY PUNTOS :( LKDSAFLÑDKMDFÑVDV

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

gracias por los puntos que tengas un buen día :)

Explanation:

<em>-ANGIEEE</em>

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A certain system absorbs 350 joules of heat and has 230 joules of work done on it what is the value of
marysya [2.9K]
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate). 

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 

(10.0 g) / (0.0480 mol) = 208.3 g/mol 

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
7 0
3 years ago
I need help due today!!!!!
Hitman42 [59]

Answer:

A. 5 g/mL

Explanation:

5 0
3 years ago
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Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c
garri49 [273]
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
                             moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
                                                    = 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane. 
5 0
3 years ago
Read 2 more answers
Chlorine pentafluoride gas is collected at -17.0 °C in an evacuated flask with measured volume of 35.0 L. When all the gas has b
Svet_ta [14]

Answer:

1. The mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of Chlorine pentafluoride, ClF5, is 0.3mole

Explanation:

1. To solve the mass of Chlorine pentafluoride, ClF5, first, let us calculate the molar mass of ClF5. This is illustrated below:

Molar Mass of ClF5 = 35.5 + (5 x 19) = 35.5 + 95 = 130.5g/mol

From the ideal gas equation:

PV = nRT (1)

Recall:

Number of mole(n) = mass (m) /Molar Mass(M)

n = m/M

Now substituting the value of n in equation 1, we have:

PV = nRT

PV = mRT/M

Now we can obtain the mass of Chlorine pentafluoride ClF5 as follow:

Data obtained from the question include:

T (temperature) = -17.0 °C = - 17 + 273 = 256K

V (volume) = 35L

P (pressure) = 0.180 atm

R (gas constant) = 0.082atm.L/Kmol

m (mass of Chlorine pentafluoride) =?

M (molar mass of Chlorine pentafluoride) = 130.5g/mol

PV = mRT/M

0.180 x 35 = m x 0.082 x 256/ 130.5

Cross multiply to express in linear form as shown below:

m x 0.082 x 256 = 0.180x35x130.5

Divide both side by 0.082 x 256

m = (0.180x35x130.5) /(0.082x256)

m = 39.16g

Therefore, the mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of ClF5 can be obtained as follow:

Mass of ClF5 = 39.16g

Molar Mass of ClF5 = 130.5g/mol

Mole of ClF5 =?

Number of mole = Mass /Molar Mass

Mole of ClF5 = 39.16/130.5g

Mole of ClF5 = 0.3mole

8 0
3 years ago
Read 2 more answers
Calculate the amount of gold deposited when a current of 5A is passed through a solution of gold salt for 2hrs 15mins. If the sa
Nataliya [291]

Answer:

a. 82.68 g b. 9.8 min

Explanation:

a. The amount of gold deposited by 5 A current in 2 hrs 15 mins

Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s

Q = 5 A × 8100 s

= 40500 C

Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C

n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole

Now n = m/M where m = mass of gold and M = molar mass of gold = 197

m = nM

= 0.42 × 197 g

= 82.68 g

b. The time taken for 6g of gold to be deposited.

We first find the number of moles of gold in 6g of gold

Since n = m/M and m = 6 g

n = 6/197 = 0.0305 mole

Q = It = nF

t = nF/I

= 0.0305 mol × 96500 C/5 A

= 2939.09 mol C

= 587.82 s

Changing t to minutes

587.82/60 s = 9.8 min

8 0
3 years ago
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