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Hoochie [10]
3 years ago
14

Which element in group 1 has the greatest tendency to lose an electron?

Chemistry
2 answers:
77julia77 [94]3 years ago
7 0
A. cesium. bacause as the elements go down a group of the atoms, the sizes of the atom will increase . we know that tge tendency to lose an electron is based on the sizes of the atoms of group 1.
melisa1 [442]3 years ago
4 0
A. cesium hope it help

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(worth 10pts!!!!!!) What did Dmitri Mendeleev contribute to science? A. He arranged the first periodic table. B. He developed th
vesna_86 [32]

Answer:

A. He arranged the first periodic table.

Explanation:

3 0
3 years ago
Read 2 more answers
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
3 years ago
Someone help1!1!1!1!!!!!
alisha [4.7K]
The question is cut off in the picture
7 0
3 years ago
Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen- containing comp
dimulka [17.4K]

Answer:

1.1 × 10² g

Explanation:

First, we will convert 1.0 L to cubic centimeters.

1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³

The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:

1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g

1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:

1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g

4 0
3 years ago
At the start of a reaction, there are 0.0249 mol N2,
gladu [14]

Answer:

Explanation:

The reaction is given as:

N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}

The reaction quotient is:

Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}

From the given information:

TO find each entity in the reaction quotient, we have:

[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}

[N_2] = \dfrac{0.024 }{3.5}

[N_2] = 0.006857

[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}

[H_2] = 9.17 \times 10^{-3}

∴

Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135

However; given that:

K_c = 1.2

By relating Q_c \ \ and  \ \ K_c, we will realize that Q_c \ \ <  \ \ K_c

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

7 0
3 years ago
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