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3 years ago

Food couldn’t reach the stomach without the esophagus and the _? please help only correct answers.

Chemistry

1 answer:

expeople1 [14]3 years ago

3 0

Food couldn’t reach the stomach without the esophagus and the throat.

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Which of the foods did you determine to be nutritious snacks?

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Fruits and vegetables

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3 years ago

How many liters of water would you need to add to 4.36 moles of NaCl to create a 9.4 M solution?

marishachu [46]

Answer:

0.464 L

Explanation:

Molarity (M) = number moles (n) ÷ volume (V)

According to the information given in this question:

number of moles (n) = 4.36 moles

Molarity = 9.4M

Volume = ?

Using M = n/V

9.4 = 4.36/V

9.4V = 4.36

V = 4.36/9.4

V = 0.464 L

Hence, 0.464L of water are needed the volume of water.

3 0

3 years ago

An example of an ____ reaction is when metals react with oxygen to form metal _____.

Reptile [31]

An example of an exothermic reaction is when metals react with oxygen to form metal Oxides

Hope this helped you- have a good day bro cya)

4 0

3 years ago

Convert 7.50 grams of glucose C6H12O6 to moles

Lera25 [3.4K]

Answer:

The number of mole is 0.04167mole

Explanation:

To convert gram to mole, we need to calculate the molecular weight of the compound

C6H12O6

C - 12

H - 1

O - 16

Molecular weight = 6 * 12 + 1 *12 + 6 * 16

= 72 + 12 + 96

= 180g/mol

To covert gram to mole

Therefore,

= 7.50g/ 180g/mol

= 0.04167 mole of glucose

3 0

3 years ago

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Answer: The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

Explanation:

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

3 0

3 years ago