All categories

3 years ago

How many calories are equivalent to 35 kilocalorie

Chemistry

2 answers:

olchik [2.2K]3 years ago

5 0

This can be very confusing.

10 kilocalories is the same as 10,000 calories so the answer is 35,000 calories.

GarryVolchara [31]3 years ago

4 0

The answer is 35,000 calories

You might be interested in

Really need help!

WARRIOR [948]

The magnitude of a star as it would appear to a hypothetical observer at a distance of 10 parsecs or 32.6 light-years. This rates how visible celestial bodies are when they are all viewed from the same distance. Luminosity: The brightness of a star in comparison with that of the sun.

7 0

3 years ago

Read 2 more answers

Calculate the number of grams of oxygen gas in a 24L container at STP

romanna [79]

There are 34 g of oxygen in the container.

We can use the Ideal Gas Law to solve this problem.

$pV = nRT$

But $n = \frac{m}{M}$ , so

$pV = \frac{m}{M}RT$ and

$m = \frac{pVM}{RT}\\$

STP is 0 °C and 1 bar, so

$m = \frac{\text{1 bar} \times \text{24 L} \times 32.00 \text{ g}\cdot\text{mol}^{-1}}{\text{0.083 14 } \text{bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1}\times\text{273.15 K} } = \textbf{34 g}\\$

7 0

3 years ago

The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. what is this distance in cen

e-lub [12.9K]

pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter. We can change the distance d = 115 pm to units of centimeters. d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m) d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm The distance in centimeters is 1.15 x 10^{-8} cm

7 0

3 years ago

Carbohydrates, more commonly known as sugars, are made up of carbon, oxygen, and hydrogen atoms. the smallest unit of a carbohyd

lukranit [14]

I believe the statement above is true. A carbohydrate is a biological molecule consisting of carbon (C), hydrogen (H) and oxygen (O) atoms, usually with a hydrogen–oxygen atom ratio of 2:1. When a carbohydrate is broken into its component sugar molecules by hydrolysis (e.g. sucrose being broken down into glucose and fructose), this is termed saccharification.

3 0

3 years ago

Please help !!

ipn [44]

Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy = $\frac{22.4L}{1mol}\times 0.545mol=12.208L$ of volume

Hence, 12.208 L of carbon dioxide will be formed.

5 0

3 years ago