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2 years ago

Helpppp me please please

Chemistry

2 answers:

kipiarov [429]2 years ago

3 0

The answer is letter d

ANEK [815]2 years ago

3 0

The answer is D. The lamp provides heat energy that causes condensation.

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How many grams are in a sample of 0.55 mol of K?

Katyanochek1 [597]

The equation you use here is
mass =moles x Mr

So:
Moles of K - 0.55mol
Mr of K - 39.1

Mass= 0.55x39.1 =21.505g

4 0

2 years ago

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago

A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC

inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K$

Putting values in above equation, we get:

$\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L$

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0

3 years ago

2) what is the mass of 6.02 x 1023 atoms of arsenic?

zzz [600]

Data:
Arsenic Molar Mass = 74,9216 ≈ 75 u (atomic mass unit)

Solving:

1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
..................................X -------------- 1 molecule

6,02*10²³X = 75
$x = \frac{75}{6,02*10^{23}}$
$\boxed{x \approx 1,24*10^{24}grams}$

6 0

2 years ago

The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released b

brilliants [131]

Combustion is a chemical reaction between a fuel and an oxidant, oxygen, to give off combustion products and heat. Complete combustion results when all of the fuel is consumed to form carbon dioxide and water, as in the case of a hydrocarbon fuel. Incomplete combustion results when insufficient oxygen reacts with the fuel, forming soot and carbon monoxide.

The complete combustion of propane proceeds through the following reaction:
$C_{3} H_{8}$ + $5O_{2}$ --> $3CO_{2}$ + $4H_{2}O$

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic.

The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.

14.50 kg propane x 1000 g x 1 mole propane x 2220 kJ 
1 kg 44.1 g 1 mole

= 729,931.97 kJ

8 0

2 years ago