All categories

2 years ago

The mass of sample A is 100.0 g, then removed it from a graduated

Chemistry

1 answer:

yan [13]2 years ago

3 0

Answer: 25g/ml

Explanation:

Density = 100.0g/ 4.0mL = 25g/mL

You might be interested in

What is the net ion charge of Beryllium?

KATRIN_1 [288]

Explanation:

Beryllium is a group 2 element and its atomic number is 4. Electronic configuration of beryllium is $1s^{2}2s^{2}$ .

Since, a beryllium contains two valence electrons so, in order to attain stability it will readily lose its 2 valence electrons.

Therefore, a beryllium atom upon losing two valence electrons will acquire a +2 charge.

Thus, we can conclude that the net ion charge of Beryllium is +2.

7 0

2 years ago

Used to measure temperature in 8th grade science.

zvonat [6]

Answer:

Fahrenheit

Explanation:

4 0

2 years ago

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0

2 years ago

I need help ASAP <br> Extra points & brainlest:)

Oksana_A [137]

Answer:

it is c

Explanation:

3 0

2 years ago

Read 2 more answers

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0

2 years ago