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Hatshy [7]
3 years ago
13

How many liters would you need to make a 2 M solution if you have 3 mol of Sodium Hydroxide

Chemistry
1 answer:
Kipish [7]3 years ago
6 0

Answer:

A 1 molar (M) solution will contain 1.0 GMW of a substance dissolved in water to make 1 liter of final solution. Hence, a 1M solution of NaCl contains 58.44 g.

Explanation:

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____________ enables you to ride a bike without skidding and falling.
Marina86 [1]
Sliding or rolling friction
4 0
2 years ago
On a phrase diagram or heating curve, the freezing point is the same as what ?
yuradex [85]

Answer:

Pelting point.

Explanation:

Freezing point is the same is the melting point.

8 0
3 years ago
A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure
GenaCL600 [577]

Answer:

P_A=4.20atm\\\\P_B=17.1atm

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

PV=nRT

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm

Best regards!

8 0
3 years ago
How many protons does the element Na have
kramer

Answer:

11 protons

Explanation:

Sodium (Na)'s atomic number or proton number is 11, which means it contains 11 protons

7 0
2 years ago
Read 2 more answers
The data below shows the change in concentration of dinitrogen pentoxide over time, at 330 K, according to the following process
tensa zangetsu [6.8K]

Answer: a) 1.7\times 10^{-4}

b) 3.4\times 10^{-4}

Explanation:

The reaction is :

2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)

Rate = Rate of disappearance of N_2O_5 = Rate of appearance of NO_2

Rate =  -\frac{d[N_2O_5]}{2dt} = \frac{d[NO_2]}{4dt}

Rate of disappearance of N_2O_5 = \frac{\text {change in concentration}}{time} = \frac{0.100-0.066}{200-0}=1.7\times 10^{-4}

a) Rate of disappearance of N_2O_5 = -\frac{d[N_2O_5]}{2dt}

Rate of appearance of NO_2 = \frac{d[NO_2]}{4dt}

b) Rate of appearance of NO_2 =  \frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}

8 0
2 years ago
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