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3 years ago

What is 0.00034 in exponential form

Chemistry

1 answer:

Verizon [17]3 years ago

7 0

Answer:

3.4 × 10 ^− 4

Explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

10 . If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

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Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11

Leona [35]

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

7 0

3 years ago

When 3-methyl-3-pentanol is treated with chromic acid, Group of answer choices

san4es73 [151]

Answer: The orange color remains unchanged. (B)

Explanation:

7 0

4 years ago

Read 2 more answers

If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?

MrMuchimi

We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.

The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.

Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, $\frac{1 X 0.75}{2}$ moles = 0.375 moles of nitrogen is required.

Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.

Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.

5 0

3 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

3 years ago

An unknown or changeable quantity is called a(n)...

alexgriva [62]

I think the answer would be dependent variable. An unknown or changeable quantity is called a dependent variable. It <span>is what you measure in the experiment and what is affected during the experiment. Hope this answers the question. Have a nice day.</span>

6 0

3 years ago