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3 years ago

The mass of an object is found by _____.

Chemistry

2 answers:

Vikki [24]3 years ago

8 0

that's the answer comparison with standard mass units

jek_recluse [69]3 years ago

7 0

<em>Hello!</em>

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comparison with standard mass units

is your answer

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Question 1 how many moles of helium are found in a balloon that contains 5.5 l of helium at a pressure of 1.15 atm and a tempera

denpristay [2]

Using pv=nRT
T= 22+273 K
P=1.15 Pa
R=8.31
V =5.5... (u did not write the unit so...)

Therefore n,mole= (1.15×5.5) ÷ (273+22)(8.31)

3 0

3 years ago

Read 2 more answers

How would I solve this kind of equation ?3.346 x 10^6km to m

mash [69]

Answer:

Explanation:

This question is asking to convert a measurement in kilometers to meters

5 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

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

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

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

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

A chemistry student in lab needs to fill a temperature-control tank with water. The tank measures 24.0 cm long by 21.0 cm wide b

Natalija [7]

Answer:

The required volume of water the student needs is 4.9 litres of water

Explanation:

From the diagram related to the question, we have;

The dimensions of the tank are;

Length of tank = 24.0 cm = 0.24 m

Width of tank = 21.0 cm = 0.21 m

Depth of tank = 13.0 cm. = 0.13 m

Allowance provided between the top of the tank and the top of the water = 2.0 cm

Diameter of the round bottom flask, D = 10.5 cm = 0.105 m

Therefore, the radius of the round bottom flask, r = 0.105/2 = 0.0525 m

Therefore we have;

Depth of water in the tank = Depth of tank - Allowance provided between the top of the tank and the top of the water

∴ Depth of water in the tank = 13.0 - 2.0 = 11.0 cm = 0.11 m

Given that the flask is immersed in the water contained in the tank to raise the tank water level, we have;

Volume of water + Volume of flask in the tank = Length of tank × Width of tank × Depth of water in the tank

Volume of water + Volume of flask in the tank = 0.24 × 0.21 × 0.11 = 0.005544 m³ = 0.005544 m³× 1000 l/m³ = 5.544 l

The volume of the spherical flask = 4/3·π·r³ = 4/3·π·0.0525³ = 6.06×10⁻⁴ m³ = 6.06×10⁻⁴ m³ × 1000 l/m³ = 0.606 l

The required volume of water the student needs , V = Volume of water + Volume of flask in the tank - The volume of the spherical flask = 5.544 l - 0.606 l = 4.9 l.

6 0

3 years ago

A 10.0-L rigid container holds 3.00 mol H2 gas at a pressure of 4.50 atm. What is the temperature of the gas?

prisoha [69]

We can approach this problem using the ideal gas law which is as follows:

PV = nRT

P = pressure
V = volume
n = number of moles
R = gas constant, 0.08206 Latm/Kmol
T = temperature

We are asked to solve for temperature and can rearrange the equation to solve for T:

PV = nRT
T = PV/nR

Now we simply plug in the data to solve for T:

T = (4.50 atm)(10.0 L)/(3.00 mol)(0.08206 Latm/Kmol)
T = 183 K

The temperature of the gas is 183 K.

3 0

3 years ago