E/St90 232 USDR11ESEnvestorite

If 50.0 grams of HCl were dissolved in enough water to make a solution with a total volume of 5.0 liters, what would be the mola

SVETLANKA909090 [29]

The molarity of a solution is found to be 10 M.

Explanation:

Molarity of a solution = moles solute

Litres solution

Therefore,M = 50

5

Molarity of a solution = 10 M

5 0

2 years ago

Material use make car tyres and chewing gum?

Ulleksa [173]

Answer:

Rubber

Make me brainlyist

3 0

3 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

Answer: The mass of original oxalic acid sample is 6.75 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

2 years ago

A(n)____ is a ripened ovary

V125BC [204]

A(n) fruit is a ripened ovary

7 0

2 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

kaheart [24]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

6 0

2 years ago