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Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

$= 8.16 \times 10^{ - 6} \: kg$

$\boxed{density = \frac{mass}{volume} }$

Density of iron bar

$= \frac{0.0642}{8.16 \times 10^{ - 6} }$

= 7870 kg/m³ (3 s.f.)

7.

$\boxed{mass = density \: \times volume}$

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

$= 12 \times 10^{ - 6} \: m^{3} \\ = 1.2 \times 10^{ - 5} \: m ^{3}$

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

$= \frac{0.151} {1.25 \times 10^{ - 5} } \\ = 12600 \: kg/ {m}^{3}$

(3 s.f.)

6 0

3 years ago

Given the atomic mass of select elements, calculate the molar mass of each salt.

arlik [135]

Answer:

See explanation

Explanation:

Molar mass is obtained as the sum of relative atomic masses as follow;

For CaBr = 40.08 + 79.90 = 119.98 g/mol

For BeBr = 9.012 + 79.90 = 88.912 g/mol

For CdBr2 = 112.41 + 2(79.90) = 272.21 g/mol

For CuBr2 = 63.55 + 2(79.90) = 223.35 g/mol

3 0

3 years ago

Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M

kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0

3 years ago

1. What three types of atoms are in a glucose molecule?

Svetach [21]

Answer:

Carbohydrates,Monosaccharides,and disaccharides

Explanation: I'm good at science

3 0

3 years ago

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p

Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0

3 years ago

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