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3 years ago

12

I WILL RATE BRAINLIEST!!

2 answers:

aivan3 [116]3 years ago

8 0

80 protons + 120 neutrons = 200
You're right. The mass number is 200.

Taya2010 [7]3 years ago

3 0

The answer is 200 for sure!!!!!

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How can phisical change what a subtance?

solmaris [256]

Physical change can change a substance by affecting the form of a chemical change.
Examples of physical changes:
Something cut (Paper)
Molded (Bread)
Boiled (Water)
Mixed (Berries and strawberries)
HOPE THIS HELPS YOU! ^_^

3 0

3 years ago

In an experiment magnesium ribbon was heated in air. The product was found to be heavier than the original ribbon. Potassium man

KatRina [158]

Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

<h3>Addition and decomposition reactions</h3>

Magnesium burns in air to produce magnesium oxide as follows:

$2Mg + O_2 --- > 2MgO$

Potassium manganate 7 burns to produce multiple products as follows:

$2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2$

Thus, the MgO will be heavier than Mg. On the other hand, $MnO_2$ will be less heavy than $KMnO_4$ .

More on reactions can be found here: brainly.com/question/17434463

#SPJ1

7 0

2 years ago

DIA [1.3K]

V_1=49.8mL
T_1=18°C=291K
T_2=83°C=356K

Using Charles law

$\\ \sf\longmapsto V_1T_2=V_2T_1$

$\\ \sf\longmapsto V_2=V_1T_2\div T_1$

$\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}$

$\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}$

$\\ \sf\longmapsto V_2=60.9mL$

5 0

3 years ago

Can someone please answer the question, <br>What is heat?

Levart [38]

Answer:

hot

Explanation:

6 0

3 years ago

Read 2 more answers

If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions

dimulka [17.4K]

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

$P_2$ = final pressure of dry gas at STP = 760 mm Hg

$V_1$ = initial volume of dry gas = 85.0 mL

$V_2$ = final volume of dry gas at STP = ?

$T_1$ = initial temperature of dry gas = $20^oC=273+20=293K$

$T_2$ = final temperature of dry gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final volume of wet gas at STP

$\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}$

$V_2=77.4mL$

Volume of dry gas at STP is 77.4 mL.

5 0

3 years ago