Which particle has two neutrons? o 1 0 n O 2 1 H o 4 2 He O 1 1 H

What's the atomic weight of salt ?

marta [7]

Answer:

58.44 g/mol

Explanation:

58.44 g/mol

4 0

2 years ago

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Due Tomorrow! What is [OH-] of a solution with a pH of -1.0? Answer in M

Margaret [11]

M IS A LETTER but N is a letter

8 0

2 years ago

A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the

SOVA2 [1]

Answer:

94.2 g/mol

Explanation:

Ideal Gases Law can useful to solve this

P . V = n . R . T

We need to make some conversions

740 Torr . 1 atm/ 760 Torr = 0.974 atm

100°C + 273 = 373K

Let's replace the values

0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K

n will determine the number of moles

(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)

n = 0.032 moles

This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?

Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol

3 0

3 years ago

HELP HELP HELP HOW MANY GRAMS OF SUGAR CAN DISSOLVE IN 100 GRAMS OF WATER HEATED TO 40 CELCIOUS 60 POINTS OFFERED

Colt1911 [192]

your answer would be C. 240g

3 0

3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago