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2 years ago

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Although a scientific theory is well supported and widely accepted, what might cause it to change? HELPPP

1 answer:

ASHA 777 [7]2 years ago

8 0

Although a scientific theories will support and wildly excepted, what might cause of the change?

the answer would be A, New Evidence.

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How many mols of a 3M NaOH solution are in 1L of the solution?

Alex

To calculate this, we need the Molarity formula. This formula tell us that Molarity, which is a concentration unit, is equal to the number of moles divided by the volume. In this question we already have the Molarity and the Volume, so let's build our equation:

C = n/V (You can see Molarity with the letter "C" because it means concentration)

3 = n/1

n = 1 * 3

n = 3 moles of NaOH

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10 months ago

What pressure will be exerted by 0.450 mol of a gas at 25°C if it is contained in a 0.650-L vessel?

iVinArrow [24]

Hey there!

The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.

Hope this helps!

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2 years ago

How many copper atoms are in 70g of copper

Elanso [62]

Answer:

$n = 6.634\times 10^{23}\,atoms$

Explanation:

The total amount of atoms is found by multiplying Avogradro's Number and the number of moles:

$n = \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)\cdot \left(\frac{70\,g}{63.546\,\frac{g}{mol} } \right)$

$n = 6.634\times 10^{23}\,atoms$

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2 years ago

Determine how many moles of CO2 are required to produce 11.0 mol of glucose,

drek231 [11]

Answer:

Determine how many moles of CO2 are required to produce 11.0 mol of glucose,

i need points thanks for CO2moles

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2 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago