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Mariulka [41]
3 years ago
10

URGENT PLZ HELP AHHHHHHH

Chemistry
1 answer:
dexar [7]3 years ago
8 0

A, B, and D. You already answered the first one, so I wouldn't think this would be too hard

:)

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A can of butane with a pressure of 2 atm at 283k has its pressure increased to a new pressure of 5 atm. What is the temperature
Ber [7]

Answer:

New temperature T2 = 707.5 K (Approx.)

Explanation:

Given:

Old pressure P1 = 2 atm

Old temperature T1 = 283 K

New Pressure P2 = 5 atm

Find:

New temperature T2

Computation:

Using Gay-Lussac law;

P1 / T1 = P2 / T2

So,

2 / 283 = 5 / T2

New temperature T2 = 707.5 K (Approx.)

6 0
3 years ago
31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams
Alexxandr [17]
Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate things. We do as follows:

</span>Heat gained = Heat lost
mC(T2-T1) = - mC(T2-T1) 
31.5C (102.4 - 32.5) = 103.5(4.18)(32.5 - 24.5)
C = 1.57 J/C-g

Hope this helps.
5 0
3 years ago
En un experimento hacemos reaccionar 12 g de carbono con 32 g de oxígeno para formar dióxido de carbono. Razona si podemos saber
max2010maxim [7]

Answer:

La masa de óxido de carbono iv formado es 44 g.

Explanation:

En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.

En primer lugar, necesitamos escribir una ecuación química equilibrada.

C + O2 → CO2

De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.

Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?

Como tenemos una relación molar de 1: 1 en todo momento, lo que esto significa es que la masa de óxido de carbono iv producida también es la misma que la masa molar de óxido de carbono iv.

Por lo tanto, procedemos a calcular la masa molar de óxido de carbono iv Esto es igual a 12 + 2 (16) = 12 + 32 = 44 g Por lo tanto, la masa de óxido de carbono iv formado es 44 g

5 0
3 years ago
A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0
3 years ago
Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrog
goldfiish [28.3K]

Answer:

The structures are shown in the figure.

Explanation:

The primary hydrogens are those which are attached to primary carbon.

Primary carbons are the carbons which are attached to only one carbon.

Primary carbons is bonded to three hydrogens.

In order to draw such structure we will draw structures which will have carbon with three hydrogens or no hydrogens (quaternary)

The structures are shown in the figure with clear marking.

7 0
3 years ago
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