Pls help with these two questions~-~

Chemistry

1 answer:

RSB [31]3 years ago

4 0

The answer is geometry and geography you add math subtract read and follow instructions

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Name the compounds that are likely to form: CaKr, Na2S, BaCl3, MgF

mafiozo [28]

Answer:

CaKr Na2S BaCl3 MgF. CaKr-Kr is a noble gas Na2S-not balanced BaCl3-not balanced MgF-not balanced.

Explanation:

4 0

3 years ago

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm $^{3}$ or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then $\frac{19*1000}{550}$ will saturate 1L

= 34.545g will saturate 1Litre

The solubility thus is 34.55g/L

5 0

3 years ago

4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an averag

PtichkaEL [24]

Answer:

C. 0.191 M

Explanation:

Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:

$HCl~+~NaOH~->~NaCl~+~H_2O$

The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:

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

$0.100~M=\frac{mol}{0.0382~L}$

$mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH$

Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:

$0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl$