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PilotLPTM [1.2K]

3 years ago

13

Imagine you have a friend over for lunch. You decide to dissolve crackers in your soup as fast as possible. You must first incre

ase the surface area of the crackers.
how would you do it?

A. put crackers in the bottom of the bowl

B. pile the crackers on top of each other

C. break the crackers into pieces

D. wrap the crackers in plastic to keep them fresh

please answer quick UwU

2 answers:

amm18123 years ago

5 0

Answer:

C.break the crackers into pieces

Gelneren [198K]3 years ago

4 0

Answer:

C. break the crackers into pieces

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What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

tester [92]

Answer:

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] <em>(1)</em>

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74 log₁₀ [A⁻] / 4,00 moles

-0,24 = log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

<em>2,30 moles = [A⁻] </em>

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) =<em> 156,4 g of sodium formate</em>.

I hope it helps!

7 0

3 years ago

How many covalent bonds are there in one molecule of carbon dioxide, CO2?

Nutka1998 [239]

It would be two because there’s 1 Carbon and 2 oxygen

6 0

3 years ago

Read 2 more answers

I need help with this question please someone tell me the answer for this

Andrej [43]

Aueyeowosy yo this is my answer

5 0

2 years ago

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

Wavelength of the photon emitted = $\lambda =646.3 nm =646.3\times 10^{-9} m$

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

$E=\frac{hc}{\lambda }$

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

3 0

3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

<u>Step 2:</u> 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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago