I need help with this worksheet that's my homework, I would need help to do it all

How many liters of 1.75 M solution could be made using 35 grams of NaCl?

dolphi86 [110]

Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
$M = \frac{m}{MM*V}$

Solving:
$M = \frac{m}{MM*V}$
$1.75 = \frac{35}{58.44*V}$
$1.75*58.44V = 35$
$102.27V = 35$
$V = \frac{35}{102.27}$
$\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Which chemical equation best represents the law of conversation of matter?

dem82 [27]

H+O>h2o I believe this is the answer

8 0

3 years ago

Read 2 more answers

Discribe how continents and oceans are carried on tectonic plates

Vanyuwa [196]

This question seems to be very basic .the ocean is above the plate, the plate material is heavier than water (it being rock). That's pretty much all of it. The plates grind a little and new land pushes up at plate boundaries but this does not seem to be related. The heavier material is below and the lighter above, those being rock and water respectively.
hope this helped u

7 0

2 years ago

A chemist adds 0.50L of a 0.485 M copper(II) sulfate CuSO4 solution to a reaction flask. Calculate the millimoles of copper(II)

den301095 [7]

Explanation:

It is given that volume is 0.50 L and molarity is 0.485 M. Hence, number of millimoles will be calculated as follows.

Number of millimoles = Molarity × Volume

As there are 1000 mL in 1 L. So, 0.50 L equals 500 mL.

Therefore, putting the given values into the above formula as follows.

Number of millimoles = Molarity × Volume

= 0.485 M × 500 mL

= 242.5

Thus, we can conclude that 242.5 millimoles of copper(II) sulfate has been added by the chemist to the flask.

4 0

3 years ago

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

6 0

3 years ago