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1 year ago

Sound _____.

Chemistry

2 answers:

zlopas [31]1 year ago

5 0

The answer is the first one that say Travels in longitudinal waves

valkas [14]1 year ago

3 0

It travels in longitudinal waves.

You might be interested in

What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric

Gre4nikov [31]

The mass of NaCl formed is 8.307 grams

calculation

step 1: write the equation for reaction

Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O

Step 2: find the moles of Na₂CO₃

moles = mass/molar mass

The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles = 7.5 g/106 g/mol =0.071 moles

Step 3: use the mole ratio to determine the mole of NaCl

Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles

Step 4: calculate mass of NaCl

mass= moles x molar mass

the molar mass of NaCl= 23 +35.5 =58.5 g/mol

mass = 0.142 moles x 58.5 g/mol =8.307 grams

8 0

1 year ago

Read 2 more answers

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou

____ [38]

Answer:

$m_{Mg}=30.8mgMg$

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

$n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2$

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

$m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg$

Best regards.

3 0

1 year ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

1 year ago

How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Ann [662]

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0

1 year ago

Determine the type of each chemical equation describing a precipitation reaction. C a 2 + ( a q ) + S O 4 2 − ( a q ) ⟶ C a S O

Zepler [3.9K]

Answer:

C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )

Explanation:

A precipitation reaction is a type of displacement reaction which a precipitate forms. The precipitate would be in the solid state, different from the other products so it can be separated or removed from the reaction.

C a 2 + ( a q ) + S O 4 2 − ( a q ) ⟶ C a S O 4 ( s )

This is wrong because C a S O 4 is the the only product formed.

C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )

This is the correct option, The precipitate is C a S O 4.

C a 2 + ( a q ) + 2 B r − ( a q ) + 2 N a + ( a q ) + S O 4 2 − ( a q ) ⟶ 2 N a + ( a q ) + 2 B r − ( a q ) + C a S O 4 ( s )

This is the ionic equation for the precipitation reaction

8 0

1 year ago