All categories

3 years ago

Help me plz, I need help on this.

Chemistry

1 answer:

kirza4 [7]3 years ago

4 0

Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

You might be interested in

If your really kind please help I just want help

Eva8 [605]

Thyroid hormones T4, T3
Cortisol
Estrogen or testosterone
Insulinlike growth factor-I (IGF-I)

4 0

3 years ago

How do I find the water vapor density

ludmilkaskok [199]

answer : D= m/v

Explanation:

all about density and mass and size of the molecules in a liquid and how close they are packed together

5 0

3 years ago

1. Which of the following is NOT a cation?

Bezzdna [24]

Hey Madoudou

The correct answer is option B (sulfate)

The reason is because "Sulfate" has a negative sign.

In order for it to be a cation, it must have positive sign such as "iron(lll)ion

I hope this helps~

3 0

3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

3 years ago