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3 years ago

Will mark brainliest if correct!!

Chemistry

1 answer:

german3 years ago

4 0

Answer:

Na2O

Explanation:

it is because oxygen has a lower valence compared to sodium

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Susie cooked sausages on a barbecue. Fat and water in the sausages changed state. which two changes of state occur during this p

padilas [110]

Answer:

The options a and b are correct

Explanation:

This options provided to the question are the answers to the question. But for clarification. When the fat melts, the change of state that occurs here is from solid to liquid (which is called melting) while the change of state that occurs when the water evaporated is from liquid to gas (which is called evaporation).

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2 years ago

Methane gas is produced from the reaction of solid carbon and hydrogen gas: C(s)+2H2(g)→CH4(g) . How many liters of hydrogen gas

Stels [109]

Answer:

80 liters

Explanation:

At STP, 1 mole of ideal gas has a volume of 22.4 liters.

Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.

40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂

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3 years ago

In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, how many moles of Mg(OH)2 are produced for every mole of MgCl2?. A. 3. B. 1. C. 2.

aksik [14]

Mole ratio:

 MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

1 mole MgCl₂ -------------- 1 mole Mg(OH)₂

Answer B

hope this helps!

4 0

3 years ago

Read 2 more answers

2.5 gm of a sample of Nahco3 when strongly.

mr_godi [17]

Answer:

Sry but i hope u will get the right answer and have a great christmas only 3 more days!

Explanation:

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3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$