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2 years ago

On the moon, there is no air resistance. What forces (F) are acting on the hammer and

Chemistry

2 answers:

spayn [35]2 years ago

6 0

Answer:

gravitational force

Explanation:

hope it helps bud

Kipish [7]2 years ago

3 0

Answer:

Gravitional Force

Explanation:

Every planet has a gravitational pull which attracts all objects; the moon is no exception. It says forces in the question but I know for sure gravitational force is apart of it.

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The CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If the

aleksley [76]

Answer:

It will require second round of the cycle to release $14C0_2$

Explanation:

Reason behind the requirement of second round of the cycle to release $CO_2$ -:

The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.

7 0

1 year ago

Write the balanced equation for the reaction of aqueous sodium carbonate and aqueous calcium chloride dihydrate to form solid ca

faltersainse [42]

Na2CO3 (aq) + CaCl2H4O2 (aq) = CaCO3 (s) + 2 NaCl (aq) + 2 H2O (l)

5 0

2 years ago

Read 2 more answers

In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic

Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

$\bar \nu = \dfrac{1}{\lambda}$

Thus, if λ = 3 µm,

$\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}$

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

$\bar \nu = \mathbf{\dfrac{f}{c}}$

Thus, if f = 90 THz

$\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}$

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0

2 years ago

What volume (in liters) of a solution contains 0.14 mol of KCl?

oksano4ka [1.4K]

Answer:

$\boxed {\boxed {\sf 0.078 \ L }}$

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

moles of solute = 0.14 mol KCl
molarity= 1.8 mol KCl/ L
liters of solution=x

Substitute these values/variables into the formula.

$1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}$

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}$

$1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl$

$1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl$

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

$\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

$x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

The units of moles of potassium chloride cancel.

$x= \frac{0.14 }{1.8 L}$

$x=0.07777777778 \ L$

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

$x \approx 0.078 \ L$

There are approximately 0.078 liters of solution.

5 0

2 years ago

A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a

Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0

2 years ago