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3 years ago

HELP ASAP!!

Chemistry

2 answers:

swat323 years ago

8 0

Answer:

B and C

Explanation:

edge 2021

I just did it

bixtya [17]3 years ago

5 0

Answer:

Emission of light

color change:)

Explanation:

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Which of the following is a neutralization reaction?

MrRa [10]

Answer:

Explanation:

B, Hcl + NaoH = NaCl + H2O

6 0

3 years ago

A cubic object has a volume of 250 cm. What is its density if its mass is 5 points<br> 257.3 g?

ira [324]

Answer:

<h3>The answer is 1.03 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 257.3 g

volume = 250 cm³

We have

$density = \frac{257.3}{250} \\ = 1.0292$

We have the final answer as

Hope this helps you

8 0

3 years ago

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

gogolik [260]

<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.

<u>Explanation:</u>

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1): $H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)$

(2): $H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)$

(3): $H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)$

(4): $H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)$

Calculating the heat absorbed for the process having the same temperature:

$q=m\times \Delta H_{(f , v)}$ ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and $\Delta H_{(f , v)}$ is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

$q=m\times C_{s,l}\times (T_2-T_1)$ ......(ii)

where,

$C_{s,l}$ = specific heat of solid or liquid

$T_2\text{ and }T_1$ are final and initial temperatures respectively

<u>For process 1:</u>

We are given:

$m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC$

Putting values in equation (i), we get:

$q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J$

<u>For process 2:</u>

We are given:

$m=46g\\\Delta H_{fusion}=334J/g$

Putting values in equation (i), we get:

$q_2=46g\times 334J/g\\\\q_2=15364J$

<u>For process 3:</u>

We are given:

$m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC$

Putting values in equation (i), we get:

$q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J$

<u>For process 4:</u>

We are given:

$m=46g\\\Delta H_{vap}=2260J/g$

Putting values in equation (i), we get:

$q_4=46g\times 2260J/g\\\\q_4=103960J$

Calculating the total amount of heat released:

$Q=q_1+q_2+q_3+q_4$

$Q=[(2424.2)+(15364)+(19255.6)+(103960)]$

$Q=141003.8J=141.004kJ$ (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

7 0

3 years ago

Give the word and chemical equations (including state symbols) for the reaction of potassium and water

zepelin [54]

Answer:

Zinc is Zn and Magnesium is Mg

7 0

3 years ago

If the pressure of 50.0 mL of oxygen gas at 100°C increases from 735 mm Hg to 925 mm Hg, what is

laila [671]

Answer: .039L

Explanation:

8 0

3 years ago