All categories

3 years ago

What will happen to an object

Chemistry

2 answers:

Natali5045456 [20]3 years ago

7 0

Answer:

Explanation:

For Connexus/Edge

quester [9]3 years ago

3 0

If an object has a higher density than the fluid it is in (fluid can mean liquid or gas), it will sink. If it has a lower density, it will float. Density is determined by an object's mass and volume. If two objects take up the same volume, but have one has more mass, then it also has a higher density.

You might be interested in

The metric system is only used in France? True or false

Olegator [25]

False it’s used in a lot of other countries

8 0

4 years ago

Read 2 more answers

how many liters of O2 are needed to produce 5.62 g of SO2 in the following reaction? 4FeS2+11O2 ->2Fe2O3+8SO2

Liula [17]

8moles of SO_2 need 11mol O_2

1mol need=11/8=1.4molO_2

Moles of sO_2

$\\ \rm\hookrightarrow \dfrac{5.62}{64}=0.1$

Moles of O_2

1.4(0.1)=0.14mol

We know

1mol at STP=22.4L
1.4mol=0.14(22.4)=3.13L O_2

5 0

3 years ago

Help! With 22! And 21! I’ll give brainliest. ASAP!

Jet001 [13]

Answer:

21. c

22. Water is pulled by gravity down below the cave mud. Go to GROUNDWATER. Water gathers into a passage as a spring & flows out

Explanation:

7 0

3 years ago

Read 2 more answers

Write a balanced half-reaction for the oxidation of solid manganese dioxide to permanganate ion in basic aqueous solution. Be su

AlexFokin [52]

Answer:

8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻

Explanation:

Let's consider the following oxidation half-reaction that takes place in basic aqueous solution.

Mn(s) ⇒ MnO₄⁻(aq)

First, we will perform the mass balance. We will add 4 H₂O to the products side and 8 OH⁻ to the reactants side.

8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l)

Finally, we will perform the charge balance by adding 7 electrons to the products side.

8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻

3 0

3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago