1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lilavasa [31]
2 years ago
11

Need answer now! 100 points

Chemistry
2 answers:
ratelena [41]2 years ago
5 0

Answer: D

divide the moles of the substances given in the question by the co-efficient in the equation. Smallest quotient is the limiting reactant

s344n2d4d5 [400]2 years ago
3 0

Answer:

D) HBr

Explanation:

Given data:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

Limiting reactant = ?

Solution:

2Al + 6HBr      →      2AlBr₃ + 3H₂

now we will compare the moles of reactants with products.

                Al          :         AlBr₃

                   2         :            2

                    8         :          8

                 Al           :          H₂

                 2             :          3

                 8             :           3/2×8 = 12

                 HBr         :         AlBr₃

                   6           :              2

                    8           :            2/6×8 = 2.67

                 HBr         :            H₂

                   6            :            3

                   8             :           3/6×8 = 4

HBr produced less number of moles of product thus it will act as limiting reactant.

You might be interested in
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0
3 years ago
An ideal gas is made up of gas particles that
Hatshy [7]

Answer:

no it is mad out of gas

Explanation:

4 0
3 years ago
Read 2 more answers
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
Write balanced equations for each of the processes described below. (Use the lowest possible coefficients. Omit states-of-matter
Aleks [24]

Answer:

d

Explanation:

7 0
3 years ago
A low energy electromagnetic wave would also have?
swat32
The answer is low frequency and long wavelength
8 0
3 years ago
Other questions:
  • How objects with different masses and volumes have the same density?​
    12·1 answer
  • If 1 gram of calcium chloride is dissolved in 100 mL of pure water, what temperature change will be observed
    12·1 answer
  • A loss of electrons by an atom creates an anion? <br> true or false
    15·2 answers
  • How would the melting point of an ice cube compare to the melting point of a large ice sculpture?
    11·2 answers
  • Please help me I will give you the brain thing and extra points. 3/9<br> image below
    5·2 answers
  • Net ionic equation for 2AgF(aq) + k2S = Ag2S (s) + 2KF(aq)
    13·1 answer
  • Balence F2+NaI. what arw the coefficients and is it an anion or a cation?​
    10·1 answer
  • A. Use the words below to complete the sentences.
    15·2 answers
  • HELP MEEEE FINALS THIS WEEK!!! I’ve gotten those so far I just need the other ones!!
    15·1 answer
  • The molecular formula<br>of oxide of silicon​
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!