Write the scientific term . the metallic oxides which desolve in water (...............)

How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?

cestrela7 [59]

In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

H-atoms in CH3OH = 2 * 6.02 * 1023 = ~1.2 * 1024

8 0

3 years ago

What is the name of the ionic compound AlBr3?

Aleksandr [31]

Answer: Correct name will be is aluminum bromide

Explanation:

In a molecular formula , $AlBr_3$

Aluminium atoms present = 1

Bromine atoms present = 3

Charge on aluminium is +3 and charge on bromine is -1.

While naming:

Name of the cation is written first. Simple name of the element is written
After name of cation name of an anions written with suffix 'ide' in the end.

So, the name of $AlBr_3$ will be aluminium bromide.

7 0

3 years ago

Read 2 more answers

2

son4ous [18]

Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

moles of HCl= $molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol$

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water = $volume \times density=150.0ml\times 1.0g/ml=150.0g$

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

q = heat released

m = mass = 150.0 g

c = specific heat = $4.184J/g^0C$

$\Delta T$ = change in temperature = $4.83^0C$

$q=150.0\times 4.184\times 4.83$

$q=3031.3J$

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = $\frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ$

Thus molar enthalpy change is 73.04 kJ/mol

8 0

3 years ago

2) If 100.0 J of heat are added to 20.0 g of water at 30.0°C, what will be the final

anzhelika [568]

The final temperature of the water will be 31.2 °C... i don’t know the second one sorry :(

3 0

2 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

2 years ago

Write the scientific term .the metallic oxides which desolve in water (...............)