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Sonja [21]
3 years ago
15

True or False - when setting up your conversion problems every numerical value must have a unit on it.

Chemistry
1 answer:
Anastaziya [24]3 years ago
3 0
I think its true but im not 100 percent sure 
You might be interested in
How many moles of argon are contained in 58 L of At at STP?
Harman [31]

Answer:

n = 2.58 mol

Explanation:

Given data:

Number of moles of argon = ?

Volume occupy = 58 L

Temperature = 273.15 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K

58 atm.L = n × 22.43 atm.L/ mol.

n = 58 atm.L / 22.43 atm.L/ mol

n = 2.58 mol

8 0
3 years ago
If i have 8 L of 0.25 M solution, how many miles of MgCl2 are present
qaws [65]

Answer:

2 moles

Explanation:

The following were obtained from the question:

Molarity = 0.25 M

Volume = 8L

Mole =?

Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

Mole of MgCl2 = 2 moles

Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution

4 0
3 years ago
The rate of decomposition of acetaldehyde, CH_3 CHO(g), into CH_4(g) and CO(g) in the presence of I_2(g) at 800 K follows the ra
FrozenT [24]

Answer:

Explanation:

Catalyst is I2 . Because I2 is reacted with starting material in step 1 and generated in second step

Rate limiting step is step 1. Because in rate equation CH3CHO and I2 is mentioned. Hence the overall rate of reaction is depending CH3CHO and I2. Rate limiting step is step 1

5 0
3 years ago
The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample
IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

             Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

                Q = 0.031 calories/gram °C x 10.0 g x 20 °C

                Q = 6.2 calories

So option A is correct = 6.2 calories

6 0
3 years ago
Lab report calorimetry and specific heat <br> HELP PLZ
dolphi86 [110]

Explanation:

Specific heat capacity can be calculated using the following equation: q = mc∆T In the equation q represents the amount of heat energy gained or lost (in joules ), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and ∆T is the temperature change of the substance 

8 0
3 years ago
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