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beks73 [17]
2 years ago
10

Kia is doing an experiment in science lab. She is given a beaker containing 100 g of liquid. The beaker has markings on the side

for measuring volume. The water comes up to the 100 mL mark. Kia puts the liquid on a hot plate by mistake. By the time she realizes the mistake, half of her liquid has evaporated.
Kia still needs 100 g of liquid for her experiment. To find out how much liquid she has to replace, she needs to re-weigh her liquid. However, the balance she used before is now broken. The teacher tells Kia that she can tell how much water is left by looking. The water now comes up to the 50 mL mark.

How much mass does Kia's remaining water have?
Chemistry
2 answers:
Semmy [17]2 years ago
7 0
Kia's remaining water has a mass of 50g. You can set it up as a proportion knowing that 100ml of water has a mass of 100g and thus 50ml of water would weight 50g
tiny-mole [99]2 years ago
3 0

the correct answer is 50g

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All the elements in period 3 have this characteristic
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What is produced during the replacement reaction of ba(no3)2 and na2so4? 2bana 2no3so4 2nano3 baso4 nano3 baso4 bana2 (no3)2so4
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2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Procedures involved:

The cations or anions may transfer positions in this twofold replacement/displacement reaction, which results in AB + CD AD + CB. In such a reaction, water, an insoluble gas, or an insoluble solid must be one of the byproducts (precipitate). The reaction in question has the following molecular equation:

2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Double displacement:

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1 year ago
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7 0
3 years ago
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2How many grams of aluminum sulfate would be formed if 250g H2SO4 completely reacted with alumi
bazaltina [42]

Answer:

290.82g

Explanation:

The equation for the reaction is given below:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 54 + 288 = 342g

Now, we can obtain the mass of aluminium sulphate formed by doing the following:

From the equation above:

294g of H2SO4 produced 342g of Al2(SO4)3.

Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3

Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.

7 0
3 years ago
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