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
Crank
3 years ago
5

What 2 types of rock make up a large majority of the composition of earth crust?

Chemistry
1 answer:
maxonik [38]3 years ago
6 0
Igneous and <span>granite are the rocks.</span>
You might be interested in
mage result for alz... Warm-Up Exercises for Chapter 2 - Acid-Base Reactions roblem 2.3 H2O is the acid and NH2 is the base. Exp
laila [671]
Type this question into apex learning! Gives you answer and step by step explanation
5 0
3 years ago
What is true of a basic solution at room temperature? it has a ph value below 7. it has a greater concentration of hydroxide com
slega [8]
The true statement about basic solution at room temperature is that it has a greater concentration of hydroxide compared to hydronium ions.
Basic solutions have always pH greater than 7.
Basic solutions have bitter and caustic taste.
Basic solutions are not used as conductors in car batteries, acidic electrolytes are used in car batteries.
7 0
3 years ago
A sample of ideal gas is in a sealed container. The pressure of the gas is 505torr, and the temperature is 43 degree celsius. If
Contact [7]
Use PV=nRT to find V assuming n is one and R= 8.31 then use the answer to find P2
8 0
3 years ago
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
Other questions:
  • If you start with 6 mol of nitrogen gas (N2+) what mass (g) of ammonia (NH4) will be produced?
    9·1 answer
  • What is the difference between a sodium atom (Na) and a sodium ion (Na1+) ?
    6·1 answer
  • Why is the electronegativity of an element important?
    10·2 answers
  • Volume of a cone is 200.96 what is the height of the cone
    11·1 answer
  • What do non metals do when they ionize
    5·1 answer
  • THIS ANSWER CORRECT?
    10·2 answers
  • Can some one help whith these
    8·2 answers
  • WDYT a mixture is? Please help
    12·1 answer
  • How is latitude related to temperature?
    14·1 answer
  • PLEASE HELP, NO BOTS PLEASE
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!