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3 years ago

2.5 moles of sodium chloride is dissolved to make 0.050 liters of solution.

Chemistry

1 answer:

KiRa [710]3 years ago

6 0

The answer is:

the molarity = 50 moles/liters

The explanation:

when the molarity is = the number of moles / volume per liters.

and when the number of moles =2.5 moles

and the volume per liters = 0.05 L

so by substitution:

the molarity = 2.5moles/0.05L

= 50 moles /L

MARK ME BRAINLIEST PLEASE!!!!!

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a 55g block of metal has a specific heat of 0.45 J/g°C. What will be the temperature change of this metal if 450 J of heat energ

Dafna11 [192]

Answer:

18.18C

Explanation: Write whats below to show your work :D

m=55

C=0.45

Q=450zj

ch.temp=?

Q=mct

450 = (55)(.45)T

450 = 24.75

/24.75 = /24.75 (cancels out)

18.18 is the answer

Hope this helps! did it with my teacher.

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3 years ago

Acids naturally present in food are safe to eat because they usually are

tangare [24]

A.Weak

B.concentrated

C.dilute

D.strong

Answer: A - Weak

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3 years ago

Read 2 more answers

According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is A. the same, because in each case there ar

aleksklad [387]

Answer:

Option "B" is correct.

Explanation:

According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In $NH_{3}$ and $CH_{4}$ , the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.

3 0

3 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$

So,

Volume of water = mass/density

$\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$$

=4.5 L (Answer)

6 0

3 years ago