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Anestetic [448]
3 years ago
8

Pls answer

Chemistry
1 answer:
schepotkina [342]3 years ago
3 0
A:transverse waves. I hope you find this helpful
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How do the densities of the two objects compare?
Marizza181 [45]
Compare the density of the object in question to the density of water. If its density is less than water, it will float. For example, oak floats because its density is 0.7 g/cm³. If the density of an object is greater than water, it will sink.
3 0
3 years ago
The molar mass of a certain gas is 80.6 g. What is the density of the gas in g/L at STP?
vfiekz [6]
Using the ideal gas equation:
PV = nRT

Substituting n with mass / Mr
PV = mRT/Mr

Density = m/V
So rearranging:
Density = PMr/RT

P = 1 atm
R = 0.082 L atm / K mol
T = 273 K
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3 0
3 years ago
What are equation for the chemical change that produces water from two hydrogen molecules and one oxygen molecule. And the react
aliya0001 [1]

Answer:

h2+O ---> H2O

reactants: H2 & O

products: H2O

Explanation:

The simple reaction that produces a water molecule from H2 and O would be the one written above, even though there are 2 hydrogen molecules, they will form an H2 molecule rather than 2 individual H molecules (almost never seen) the reactants would be your hydrogen and oxygen molecules individually before they bond to form a molecule of water (H2O) which is the product

6 0
2 years ago
Please Help Weak Acid Base Problems!!!
DiKsa [7]

Answer:

The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the sample problem. However, the variable x will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.

Explanation:

3 0
3 years ago
Read 2 more answers
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

8 0
3 years ago
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