lighting a match results in light and heat since energy was not created by the match what transformation took place

One cup of fresh orange juice contains 115 mg of ascorbic acid (vitamin c, c6h8o6). given that one cup

kodGreya [7K]

The question is incomplete. Complete question is read as:
'One cup of fresh orange juice contains 115 mg of ascorbic acid (vitamin C, C6H8O6). Given that one cup = 218.0 mL calculate the molarity of vitamin C in organic juice.'
..........................................................................................................................
Answer:
Given: weight of solute (ascorbic acid) = 115 mg = 0.115 g
Volume of solution = 218.0 mL = 0.218 L
Molecular weight of ascorbic acid = 176.12 g/mol.

Now, Molarity = $\frac{\text{weight of solute(g)}}{\text{Molecular weight X Vol. of solution(l)}}$
= $\frac{\text{0.115}}{\text{176.12 X 0.218}}$
 = 0.002995 mol/dm3

Answer: Molarity of solution = 0.002995 mol/dm3

4 0

3 years ago

Fill in the blanks with the correct words to complete the sentence.

navik [9.2K]

Answer:

Explanation:

Salt water intrusion can cause the fresh water in wells to become contaminated with saltwater.

4 0

3 years ago

How many moles of MgS2O3 are in 223 g of the compound

Alexxandr [17]

Moles of MgS2O3 = 223/molar mass of MgS2O3

= 223/136.42
= 1.634 moles.

Hope this helps!

6 0

3 years ago

Read 2 more answers

A gas has an initial volume of 52.3 L at 273 Kelvin. What is its temperature when the volume reaches 145.7 L?

erastovalidia [21]

Answer:

$\boxed {\boxed {\sf 761 \ K}}$

Explanation:

We are asked to find the new temperature of a gas after a change in volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume is initially 52.3 liters at a temperature of 273 Kelvin.

$\frac {52.3 \ L}{273 \ K}= \frac{V_2}{T_2}$

The volume reaches 145.7 liters at an unknown temperature.

$\frac {52.3 \ L}{273 \ K}= \frac{145.7 \ L }{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$52.3 \ L * T_2 = 273 \ K * 145.7 \ L$

Now the variable is being multiplied by 52.3 liters. The inverse of multiplication is division. Divide both sides by 52.3 L.

$\frac {52.3 \ L * T_2 }{52.3 \ L}=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

$T_2=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

The units of liters cancel.

$T_2=\frac{ 273 \ K * 145.7 }{52.3 }$

$T_2 = 760.5372849 \ K$

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we found, that is the ones place. The 5 in the tenths place tells us to round the 0 up to a 1.

$T_2 \approx 761 \ K$

When the volume reaches 145.7 liters, the temperature is 761 Kelvin.

8 0

3 years ago

1. When _____ are added or removed from an atom, _____ is formed

Kobotan [32]

Answer:

1.) protons or electrons 2.) ion

Explanation:

4 0

3 years ago